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RD Sharma Class 12 Exercise 3.7 Inverse Trigonometric Function Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 3.7 Inverse Trigonometric Function Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:28 PM IST

The RD Sharma Solutions for Class 12 Mathematics have helped thousands of students score good marks in their public exams. It is the best guide for students to sharpen their mathematical skills whenever they find the time. Unfortunately, inverse Trsigonometry is not a concept that every student in a class would find accessible. But with the help of RD Sharma Class 12th exercie 3.7, they can perform well in their examinations.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.7

Inverse Trignometric Functions exercise 3.7 question 1 (i)

Answer: \frac{\pi }{6}
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}\left(\sin \frac{\pi}{6}\right)
Explanation:
We know that the value of \sin \frac{\pi }{6} is \frac{1}{2}
By substituting this value in \sin ^{-1}\left(\sin \frac{\pi}{6}\right)
We get \sin ^{-1}\left ( \frac{1}{2} \right )
Let,
\begin{aligned} &y=\sin ^{-1}\left(\frac{1}{2}\right) \\ &\sin y=\frac{1}{2} \\ &\sin y=\left(\frac{1}{2}\right) \\ &\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \end{aligned}
The range of principal value of \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
and \sin (\frac{\pi }{6}) =\frac{1}{2}
Therefore, \sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6}
Hence, \begin{aligned} &\therefore \sin ^{-1}(\sin \theta)=\theta \text { with } \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}\left(\sin \frac{\pi}{6}\right) \text { is } \frac{\pi}{6} \end{aligned}


Inverse Trignometric Functions exercise 3.7 question 1 (ii)

Answer: -\frac{\pi }{6}
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)
Explanation:
First, we solve \left(\sin \frac{7 \pi}{6}\right)
\operatorname{Sin} \frac{7 \pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)
As we know \sin \left(\frac{\pi}{6}\right)=\frac{1}{2}
\sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right) \quad \because \sin [\pi+\theta]=\sin (-\theta)
\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)=\sin ^{-1}\left(-\frac{1}{2}\right)
Now, \text { let } y=\sin ^{-1}\left(-\frac{1}{2}\right)
\begin{aligned} &-\sin y=\frac{1}{2} \\ &-\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \\ &-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right) \end{aligned}
The range of principal value of \sin ^{-1}\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text { and } \sin \left(-\frac{\pi}{6}\right)=-\frac{1}{2}
\therefore \sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6} \quad\left\{\text { As } \operatorname{Sin} \frac{7 \pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)\right\}
As \begin{aligned} &\left[\sin ^{-1}(\sin x)\right]=x \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (iii)

Answer: \frac{\pi }{6}
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)
Explanation:
First, we solve \sin \frac{5\pi }{6}
\frac{5 \pi}{6}=\pi-\frac{\pi}{6}
Then,
\begin{aligned} &\sin \left(\pi-\frac{\pi}{6}\right)=\sin \left(\frac{\pi}{6}\right) \\ &\sin (\pi-\theta)=\sin (\theta) \end{aligned}
We know that the value of \sin \frac{\pi }{6} is \frac{1}{2}
\sin \frac{5 \pi}{6}=\frac{1}{2}
By substituting this value in \sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)
we get \sin ^{-1}\left(\frac{1}{2}\right)
Let y=\sin ^{-1}\left(\frac{1}{2}\right)
\begin{aligned} &\sin y=\frac{1}{2} \\ &\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \end{aligned}
The range of principal value of \sin ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } \sin \left(\frac{\pi}{6}\right)=\frac{1}{2}
\begin{aligned} &\sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6} \\ &\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)=\frac{\pi}{6} \quad \because\left[\sin ^{-1}(\sin \theta)=\theta\right] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (iv)

Answer:

Answer: -\frac{\pi }{7}
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)
Explanation:
First we solve \left(\sin \frac{13 \pi}{7}\right)
\frac{13 \pi}{7}=2 \pi-\frac{\pi}{7}
\sin \left(2 \pi-\frac{\pi}{7}\right)=\sin \left(-\frac{\pi}{7}\right) \quad \because[\sin (2 \pi-\theta)=\sin (-\theta)]
By substituting these value in
\sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)=\sin ^{-1}\left(\sin -\frac{\pi}{7}\right)
As \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
The range of principal value of \sin ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)=-\frac{\pi}{7}

Inverse Trignometric Functions exercise 3.7 question 1 (v)

Answer: -\frac{\pi }{8}
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given:
Explanation:
First we solve \left(\sin \frac{17 \pi}{8}\right)
\begin{aligned} &\sin \frac{17 \pi}{8}=\sin \left(2 \pi+\frac{\pi}{8}\right) \\ &\sin \left(2 \pi+\frac{\pi}{8}\right)=\sin \left(\frac{\pi}{8}\right) \quad \because[\sin (2 \pi+\theta)=\sin (\theta)] \end{aligned}
By substituting these values in \sin ^{-1}\left(\sin \frac{17 \pi}{8}\right)
we get \sin ^{-1}\left(\sin \frac{\pi}{8}\right)
\begin{gathered} \text { As } \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ \therefore \sin ^{-1}\left(\sin \frac{17 \pi}{8}\right)=\frac{\pi}{8} \end{gathered}


Inverse Trignometric Functions exercise 3.7 question 1 (vi)

Answer: -\frac{\pi }{8}
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}\left\{\left(\sin -\frac{17 \pi}{8}\right)\right\}
Explanation:
First we solve \left(\sin -\frac{17 \pi}{8}\right)
As we know, -\sin \theta=\sin (-\theta)
\begin{aligned} &\therefore\left(\sin -\frac{17 \pi}{8}\right)=-\sin \frac{17 \pi}{8} \\ &-\sin \frac{17 \pi}{8}=-\sin \left(2 \pi+\frac{\pi}{8}\right) \end{aligned}
-\sin \left(2 \pi+\frac{\pi}{8}\right)=-\sin \left(\frac{\pi}{8}\right) \quad \because[\sin (2 \pi+\theta)=\sin (\theta)]
\begin{aligned} &\therefore[-\sin (\theta)=\sin (-\theta)] \\ &-\sin \left(\frac{\pi}{8}\right)=\sin \left(-\frac{\pi}{8}\right) \end{aligned}
By substituting these values in \sin ^{-1}\left\{\left(\sin -\frac{17 \pi}{8}\right)\right\} we get,
\sin ^{-1}\left(\sin -\frac{\pi}{8}\right)
\begin{gathered} \text { As } \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ \therefore \sin ^{-1}\left(\sin -\frac{\pi}{8}\right)=-\frac{\pi}{8} \end{gathered}

Inverse Trignometric Functions exercise 3.7 question 1 (vii)
Answer:

{\pi }-3
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}\left(\sin 3\right)
Explanation:
We know that,\sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] which is approximately equals to \left [ -1.57,1.57 \right ]
But here x=3 , which do not lie on the above range.
∴ We know that \sin (\pi-x)=\sin (x)
\begin{aligned} &\sin (\pi-3)=\sin (3) \text { also } \pi-3 \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}(\sin 3)=\sin ^{-1}(\sin \pi-3) \\ &\text { Then, } \sin ^{-1}(\sin \pi-3)=\pi-3 \end{aligned}


Inverse Trignometric Functions exercise 3.7 question 1 (viii)

Answer:

\pi -4
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}(\sin 4)
Explanation:
We know that \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] which is approximately equal to [-1.57, 1.57]
But x=4 , which do not lie on the above range
We know,
\begin{aligned} &\sin (\pi-x)=\sin (x) \\ &\sin (\pi-4)=\sin (4) \text { also } \pi-4 \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}(\sin 4)=\sin ^{-1}(\sin \pi-4) \end{aligned}
Then,
\sin ^{-1}(\sin \pi-4)=\pi-4

Inverse Trignometric Functions exercise 3.7 question 1 (ix)

Answer: 12-4{\pi }
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}(\sin 12)
Explanation:
We know that \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] which is approximately equal to [-1.57, 1.57]
But x=12, which do not lie on the above range
We know \sin (2 n \pi-x)=\sin (-x)
Hence \sin (2 n \pi-12)=\sin (-12)
Here,
\begin{aligned} &n=2 \text { also } 12-4 \pi \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}(\sin 12)=12-4 \pi \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (x)

Answer: \pi -2
Hint: The principal value branch of function \sin ^{-1} is \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]
Given: \sin ^{-1}(\sin 2)
Explanation:
We know that \sin ^{-1}(\sin x)=x \text { with } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] which is approximately equal to[-1.57, 1.57]
But here x=2 , which do not lie on the above range,
we know that \sin (\pi-x)=\sin (x)
\begin{aligned} &\sin (\pi-2)=\sin (2) \text { also } \pi-2 \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}(\sin 2)=\sin ^{-1}(\sin \pi-2) \\ &\therefore \sin ^{-1}(\sin \pi-2)=\pi-2 \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 2 (i)

Answer: \frac{\pi }{4}
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}
Explanation:
We know that \cos \left(-\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right) \quad \because[\cos (-\theta)=\cos \theta]
Also know that, \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}
By substituting these values in \cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}
we get,
\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)
Let y=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)
\cos y=\frac{1}{\sqrt{2}}
Hence, range of principal value of \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}
\begin{aligned} &\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}=\frac{\pi}{4} \\ &\therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 2 (ii)

Answer: \frac{3\pi }{4}
Hint: The principal value branch of function \cos ^{-1} is [0,\pi ]
Given: \cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)
Explanation:
First we solve \cos \frac{5\pi }{4}
\begin{aligned} &\cos \frac{5 \pi}{4}=\cos \left(\pi+\frac{\pi}{4}\right) \\ &\cos \left(\pi+\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \quad \because[\cos (-\theta)=\cos \theta] \end{aligned}
As we know that \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}
\cos \left(\frac{5 \pi}{4}\right)=-\frac{1}{\sqrt{2}}
By substituting these value in \cos ^{-1}\left\{\cos \left(\frac{5 \pi}{4}\right)\right\}
we get,
\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)
Since [\cos (-\theta)=\cos \theta]
Let y=\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)
\begin{aligned} &\cos y=-\frac{1}{\sqrt{2}} \\ &-\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\ &\cos \left(\pi-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \\ &\cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}} \end{aligned}
Hence, range of principal value of \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}}
\cos ^{-1}\left\{\cos \left(\frac{5 \pi}{4}\right)\right\}=\frac{3 \pi}{4} \therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi]

Inverse Trignometric Functions exercise 3.7 question 2 (iii)

Answer: \frac{2\pi }{3}
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}\left ( \cos \frac{4\pi }{3} \right )
Explanation:
First we solve \cos \left ( \frac{4\pi }{3} \right )
\begin{aligned} &\cos \frac{4 \pi}{3}=\cos \left(\pi+\frac{\pi}{3}\right) \\ &\therefore \cos [\pi+\theta]=-\cos \theta \end{aligned}
\begin{aligned} &\cos \left(\pi+\frac{\pi}{3}\right)=-\frac{1}{2} \\ &\cos \left(\frac{4 \pi}{3}\right)=-\frac{1}{2} \end{aligned}
By substituting these value in \cos ^{-1}\left\{\cos \left(\frac{4 \pi}{3}\right)\right\}
\cos ^{-1}\left(-\frac{1}{2}\right)
Now,
Let y=\cos ^{-1}\left(-\frac{1}{2}\right)
\begin{aligned} &\cos y=-\frac{1}{2} \\ &-\cos \left(\frac{\pi}{3}\right)=\frac{1}{2} \\ &\cos \left(\pi-\frac{\pi}{3}\right)=-\frac{1}{2} \\ &\cos \left(\frac{2 \pi}{3}\right)=-\frac{1}{2} \end{aligned}
Hence, range of principal value of \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{2 \pi}{3}\right)=-\frac{1}{2}
\cos^{-1}(\cos\frac{2\pi}{3})=\frac{2\pi}{3}

Inverse Trignometric Functions exercise 3.7 question 2 (iv)

Answer: \frac{\pi }{6}
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)
Explanation:
First we solve \cos \frac{13 \pi}{6}
\begin{aligned} &\cos \frac{13 \pi}{6}=\cos \left(2 \pi+\frac{\pi}{6}\right) \\ &\therefore[\cos (2 \pi+\theta)]=\cos \theta \\ &\cos \left(2 \pi+\frac{\pi}{6}\right)=\cos \left(\frac{\pi}{6}\right) \end{aligned}
\begin{gathered} =\frac{\sqrt{3}}{2} \\ \cos \left(\frac{13 \pi}{6}\right)=\frac{\sqrt{3}}{2} \end{gathered}
By substituting these values in \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)
we get,
\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)
Now, \operatorname{let} y=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)
\begin{aligned} &\cos y=\frac{\sqrt{3}}{2} \\ &\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2} \end{aligned}
Hence, range of principal value of \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}
\begin{aligned} &\cos ^{-1}\left(\cos \frac{\pi}{6}\right)=\frac{\pi}{6} \\ &\therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi] \\ &\cos ^{-1}\left\{\cos \left(\frac{13 \pi}{6}\right)\right\}=\frac{\pi}{6} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 2 (v)

Answer: 3
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}(\cos 3)
Explanation:
We know that
\cos ^{-1}(\cos 3)=3,3 \in[0, \pi] \quad\left[\cos ^{-1}(\cos \theta)=\theta \text { if } 0 \leq \theta \leq \pi\right]
Hence, the answer is \cos ^{-1}(\cos 3)=3

Inverse Trignometric Functions exercise 3.7 question 2 (vi)

Answer: 2\pi -4
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}(\cos 4)
Explanation:
We know that \cos ^{-1}(\cos x)=x \text { if } x \in[0, \pi] \approx[0,3.14]

And herex=4 which do not lie in the above range

\begin{aligned} &\cos (2 \pi-x)=\cos (x) \\ &\cos (2 \pi-4)=\cos (4) \operatorname{so} 2 \pi-4 \text { belongs in }[0, \pi] \\ &\cos ^{-1}(\cos 4)=2 \pi-4 \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 2 (vii)

Answer: {2\pi }-5
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}(\cos 5)
Explanation:
We know that \cos ^{-1}(\cos x)=x \text { if } x \in[0, \pi] \approx[0,3.14]
And here x=5 which do not lie in the above range.
We know that \cos (2 \pi-x)=\cos (x)
\cos (2 \pi-5)=\cos (5) \operatorname{so} 2 \pi-5 \text { belongs in }[0, \pi]
Hence, \cos ^{-1}(\cos 5)=2 \pi-5

Inverse Trignometric Functions exercise 3.7 question 2 (viii)

Answer: {4\pi }-12
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \cos ^{-1}(\cos 12)
Explanation:
We know that \cos ^{-1}(\cos x)=x \text { if } x \in[0, \pi] \approx[0,3.14]
Herex=12 which do not lie in the above range.
We know that
\begin{aligned} &\cos (2 n \pi-x)=\cos (x) \\ &\cos (2 n \pi-12)=\cos 12 \end{aligned}
Here, n=2
Also, \begin{aligned} &4 \pi-12 \text { belongs in }[0, \pi] \end{aligned}
\cos ^{-1}(\cos 12)=4 \pi-12

Inverse Trignometric Functions exercise 3.7 question 3 (i)

Answer: \\ \\ \frac{\pi}{3}
Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given: \tan ^{-1}\left(\tan \frac{\pi}{3}\right)
Explanation:
As \tan ^{-1}(\tan x)=x \text { if } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
By applying this situation
we get
\tan ^{-1}\left(\tan \frac{\pi}{3}\right)=\frac{\pi}{3}
Hence, \tan ^{-1}\left(\tan \frac{\pi}{3}\right) \text { is } \frac{\pi}{3}

Inverse Trignometric Functions exercise 3.7 question 3 (ii)

Answer: -\frac{\pi }{7}
Hint: The range of principal value of \tan ^{-1} is \left [-\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}\left(\tan \frac{6 \pi}{7}\right)
Explanation:

First we solve \tan \frac{6\pi }{7}

\begin{aligned} &\tan \frac{6 \pi}{7}=\tan \left(\pi-\frac{\pi}{7}\right) \\ &\tan \left(\pi-\frac{\pi}{7}\right)=-\tan \frac{\pi}{7}\; \; \; \; \; \; \; \; \; \quad(\tan (\pi-\theta)=-\tan \theta) \end{aligned}

We know

\tan ^{-1}(\tan x)=x \text { if } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

\tan ^{-1}\left(\tan \frac{6 \pi}{7}\right)=-\frac{\pi}{7}

Inverse Trignometric Functions exercise 3.7 question 3 (iii)

Answer: \frac{\pi }{6}
Hint: The range of principal value of \tan ^{-1} is \left [-\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)
Explanation:
First we solve \tan \frac{7 \pi}{6}
\tan \frac{7 \pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)
As we know, \tan (\pi+\theta)=\tan \theta
\begin{aligned} &\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{\pi}{6} \\ &\therefore \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \end{aligned}
By substituting this value in \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) we get
\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)
Now, \text { let } \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y
\begin{aligned} &\tan y=\frac{1}{\sqrt{3}} \\ &\tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}} \end{aligned}
The range of principal value of \tan ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } \tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}
\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\frac{\pi}{6} \quad \text { As } \quad \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Inverse Trignometric Functions exercise 3.7 question 3 (iv)

Answer: \frac{\pi }{4}
Hint: The range of principal value of \tan ^{-1} is \left [-\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)
Explanation:
First we solve \tan \frac{9 \pi}{4}
\tan \frac{9 \pi}{4}=\tan \left(2 \pi+\frac{\pi}{4}\right)
\therefore \tan (2 \pi+\theta)=\tan \theta
\begin{aligned} &\tan \left(2 \pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4} \\ &\therefore \tan \frac{\pi}{4}=1 \\ &\tan \frac{9 \pi}{4}=1 \end{aligned}
By substituting this value in \tan ^{-1}\left(\tan \frac{9 \pi}{4}\right), we get
\tan ^{-1}(1)
let, \tan ^{-1}(1)=y
\begin{aligned} &\tan y=1 \\ &\tan \left(\frac{\pi}{4}\right)=1 \end{aligned}
The range of principal value of \tan ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } \tan \left(\frac{\pi}{4}\right)=1
\tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)=\frac{\pi}{4}\; \; \; \; \; \; \; \; \quad \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Inverse Trignometric Functions exercise 3.7 question 3 (v)

Answer: 1
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}(\tan 1)
Explanation:
As we know \tan ^{-1}(\tan x)=x \text { if } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
By substituting this condition in given question
\tan ^{-1}(\tan 1)=1
Hence, \tan ^{-1}(\tan 1)=1

Inverse Trignometric Functions exercise 3.7 question 3 (vi)

Answer: 2-\pi
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}(\tan 2)
Explanation:
As \tan ^{-1}(\tan x)=x \text { if } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
But here x=2 which does not belong to above range
\begin{aligned} &\tan (\pi-\theta)=-\tan (\theta) \\ &\tan (\theta-\pi)=\tan (\theta) \\ &\tan (2-\pi)=\tan (2) \end{aligned}
Now, 2-\piin the given range \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Hence, \tan ^{-1}(\tan 2)=2-\pi

Inverse Trignometric Functions exercise 3.7 question 3 (vii)

Answer: 4-\pi
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}(\tan 4)
Explanation:
As \tan ^{-1}(\tan x)=x \text { if } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
But here x=4 which does not belongs to above range
\begin{aligned} &\tan (\pi-\theta)=-\tan (\theta) \\ &\tan (\theta-\pi)=\tan \theta \\ &\tan (4-\pi)=\tan (4) \end{aligned}
Now, 4-\pi \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
Hence, \tan ^{-1}(\tan 4)=4-\pi

Inverse Trignometric Functions exercise 3.7 question 3 (viii)

Answer: 12-4\pi
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
Given: \tan ^{-1}(\tan 12)
Explanation:
As \tan ^{-1}(\tan x)=x \text { if } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
But here x=12 which does not belongs to above range
\begin{aligned} &\tan (2 n \pi-\theta)=-\tan (\theta) \\ &\tan (\theta-2 n \pi)=\tan \theta \end{aligned}
Here, n=2
\tan (12-4 \pi)=\tan (12)
Now, 12-4 \pi \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
Hence,

Inverse Trignometric Functions exercise 3.7 question 4 (i)

Answer: \frac{\pi }{3}
Hint: The range of principal value of \sec ^{-1} is \hspace{0.2cm}\forall x\in[0,\pi]\hspace{0.2cm},x\neq \frac{\pi}{2}
Given: \sec ^{-1}\left(\sec \frac{\pi}{3}\right)
Explanation:
As \sec ^{-1}(\sec x)=x \text { if } x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}
By applying this situation we get,
\sec ^{-1}\left(\sec \frac{\pi}{3}\right)=\frac{\pi}{3}
Hence, \sec ^{-1}\left(\sec \frac{\pi}{3}\right)=\frac{\pi}{3}

Inverse Trignometric Functions exercise 3.7 question 4 (ii)

Answer: \frac{2\pi }{3}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left(\sec \frac{2 \pi}{3}\right)
Explanation:
As \sec ^{-1}(\sec x)=x \text { if } x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}
By applying this situation we get,
\sec ^{-1}\left(\sec \frac{2 \pi}{3}\right)=\frac{2 \pi}{3}
Hence, \sec ^{-1}\left(\sec \frac{2 \pi}{3}\right)=\frac{2 \pi}{3}

Inverse Trignometric Functions exercise 3.7 question 4 (iii)

Answer: \frac{3\pi }{4}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left(\sec \frac{5 \pi}{4}\right)
Explanation:
First we solve \sec \frac{5 \pi}{4}
But \left(2 \pi-\frac{3 \pi}{4}\right) belongs to 3 quadrant
So, \sec \left(2 \pi-\frac{3 \pi}{4}\right)=\sec \frac{3 \pi}{4}
\begin{aligned} &\therefore \sec \frac{3 \pi}{4}=\sqrt{2} \\ &-\sec \frac{3 \pi}{4}=-\sqrt{2} \end{aligned}
By substituting these value in \sec ^{-1}\left(\sec \frac{5 \pi}{4}\right) we get,
\sec ^{-1}(-\sqrt{2})
Now, \text { let } y=\sec ^{-1}(-\sqrt{2})
\Rightarrow \quad \sec y=-\sqrt{2}
\Rightarrow \quad-\sec \left(\frac{\pi}{4}\right)=\sqrt{2}
\begin{aligned} &\Rightarrow \quad \sec \left(\pi-\frac{\pi}{4}\right) \\ &\Rightarrow \quad \sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}
The range of principal value of \sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\}
\sec ^{-1}\left(\sec \frac{3 \pi}{4}\right)=\frac{3 \pi}{4} \; \; \; \; \; \; \; \; \; \; \quad \therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}

Inverse Trignometric Functions exercise 3.7 question 4 (iv)

Answer: \frac{\pi }{3}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left(\sec \frac{7 \pi}{3}\right)
Explanation:
First we solve \sec \frac{7 \pi}{3}
\begin{aligned} &\sec \left(\frac{7 \pi}{3}\right)=\sec \left(2 \pi+\frac{\pi}{3}\right) \\ &\therefore[\sec (2 \pi+\theta)=\sec \theta] \end{aligned}
\begin{aligned} &\sec \left(2 \pi+\frac{\pi}{3}\right)=\sec \left(\frac{\pi}{3}\right) \\ &\therefore \sec \left(\frac{\pi}{3}\right)=2 \end{aligned}
By substituting these value in \sec ^{-1}\left(\sec \frac{7 \pi}{3}\right) we get,
\sec ^{-1}(2)
Now, \text { let } y=\sec ^{-1}(2)
\begin{aligned} &\sec y=2 \\ &\sec \left(\frac{\pi}{3}\right)=2 \end{aligned}
The range of principal value of \sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{\pi}{3}\right)=2
\begin{aligned} &\sec ^{-1}\left(\sec \frac{\pi}{3}\right)=\frac{\pi}{3} \\ &\sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 4 (v)

Answer: \frac{\pi }{5}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)
Explanation:
First we solve \sec \left(\frac{9 \pi}{5}\right)
\begin{aligned} &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(2 \pi-\frac{\pi}{5}\right) \\ &\therefore[\sec (2 \pi-\theta)]=\sec \theta \end{aligned}
\begin{aligned} &\sec \left(2 \pi-\frac{\pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \\ &\sec \left(\frac{9 \pi}{5}\right)=\sec \left(\frac{\pi}{5}\right) \end{aligned}
By substituting these value in \sec ^{-1}\left(\sec \frac{9 \pi}{5}\right) we get,
\begin{aligned} &\sec ^{-1}\left(\sec \frac{\pi}{5}\right) \\ &\therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \\ &\sec ^{-1}\left(\sec \frac{\pi}{5}\right)=\frac{\pi}{5} \end{aligned}
Hence, \sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)=\frac{\pi}{5}

Inverse Trignometric Functions exercise 3.7 question 4 (vi)

Answer: \frac{\pi }{3}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left\{\sec \left(-\frac{7 \pi}{3}\right)\right\}
Explanation:
First we solve \sec \left(-\frac{7 \pi}{3}\right)
As we know that \sec (-\theta)=\sec (\theta)
\begin{aligned} &\sec \left(-\frac{7 \theta}{3}\right)=\sec \left(\frac{7 \pi}{3}\right) \\ &\sec \left(\frac{7 \pi}{3}\right)=\sec \left(2 \pi+\frac{\pi}{3}\right) \end{aligned}
\begin{aligned} &\sec \left(2 \pi+\frac{\pi}{3}\right)=\sec \left(\frac{\pi}{3}\right) \quad \because[\sec (2 \pi+\theta)=\sec \theta] \\ &\sec \left(\frac{\pi}{3}\right)=2 \end{aligned}
By substituting these value in \sec ^{-1}\left\{\sec \left(-\frac{7 \pi}{3}\right)\right\} we get,
\sec ^{-1}(2)
Now, \text { let } y=\sec ^{-1}(2)
\begin{aligned} &\text { sec } y=2 \\ &\sec \left(\frac{\pi}{3}\right)=2 \end{aligned}
The range of principal value of \sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{\pi}{3}\right)=2
\begin{aligned} &\sec ^{-1}\left(\sec \frac{\pi}{3}\right)=\frac{\pi}{3} \\ &\sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \end{aligned}
Hence, \sec ^{-1}\left(\sec -\frac{7 \pi}{3}\right)=\frac{\pi}{3}

ma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 (vi)

Answer:

Answer: \frac{\pi }{3}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left\{\sec \left(-\frac{7 \pi}{3}\right)\right\}

Inverse Trignometric Functions exercise 3.7 question 4 (vii)

Answer: \frac{3\pi }{4}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left(\sec \frac{13 \pi}{4}\right)
Explanation:
First we solve \sec \left(\frac{13 \pi}{4}\right)
\sec \left(\frac{13 \pi}{4}\right)=\sec \left(4 \pi-\frac{3 \pi}{4}\right)
As we know [\sec 2 \pi-\theta]=\sec \theta
But here after subtract \frac{3\pi }{4} it comes in III quadrant where sec is negative
\begin{aligned} &\sec \left(4 \pi-\frac{3 \pi}{4}\right)=-\sec \left(\frac{3 \pi}{4}\right) \\ &\therefore \sec \frac{3 \pi}{4}=\sqrt{2} \\ &-\sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}
By substituting these values in \sec ^{-1}\left(\sec \frac{13 \pi}{4}\right) we get,
\sec ^{-1}(-\sqrt{2})
Now, \text { let } y=\sec ^{-1}(-\sqrt{2})
\begin{aligned} &\sec y=-\sqrt{2} \\ &-\sec y=\sqrt{2} \\ &-\sec \left(\frac{\pi}{4}\right)=\sqrt{2} \\ &\sec \left(\pi-\frac{\pi}{4}\right)=\sqrt{2} \\ &\sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}
The range of the principle value of \sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2}
\sec ^{-1}\left(\sec \frac{3 \pi}{4}\right)=\frac{3 \pi}{4}
\therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}

Inverse Trignometric Functions exercise 3.7 question 4 (viii)

Answer: \frac{\pi }{6}
Hint: The range of principal value of \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given: \sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)
Explanation:
First we solve \sec \frac{25 \pi}{6}
\sec \left(\frac{25 \pi}{6}\right)=\sec \left(4 \pi+\frac{\pi}{6}\right)
As we know \sec [2 \pi+\theta]=\sec \theta
\sec \left(4 \pi+\frac{\pi}{6}\right)=\sec \left(\frac{\pi}{6}\right)
As we know \sec \left(\frac{\pi}{6}\right)=\frac{2 \sqrt{3}}{3}
By substituting these value in \sec ^{-1}\left(\sec \frac{25 \pi}{6}\right) we get,
\sec ^{-1}\left(\frac{2 \sqrt{3}}{3}\right)
Now, \text { let } y=\sec ^{-1}\left(\frac{2 \sqrt{3}}{3}\right)
\begin{aligned} &\sec y=\frac{2 \sqrt{3}}{3} \\ &\sec \left(\frac{\pi}{6}\right)=\frac{2 \sqrt{3}}{3} \end{aligned}
The range of principal value of \sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{\pi}{6}\right)=\frac{2 \sqrt{3}}{3}
\begin{aligned} &\sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)=\frac{\pi}{6} \\ &\sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 5sub question (i)

Answer: \frac{\pi }{4}
Hint: The range of principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Given: \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{\pi}{4}\right)
Explanation:
As we know,
\operatorname{cosec}^{-1}(\operatorname{cosec} x)=x \text { when } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
By applying this situation we get,
\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{\pi}{4}\right)=\frac{\pi}{4}

Inverse Trignometric Functions exercise 3.7 question 5 (ii)

Answer: \frac{\pi }{4}
Hint: The range of principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Given: \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right)
Explanation:
First we solve \operatorname{cosec}\left(\frac{3 \pi}{4}\right)
\operatorname{cosec}\left(\frac{3 \pi}{4}\right)=\operatorname{cosec}\left(\pi-\frac{\pi}{4}\right)
As we know, [\operatorname{cosec}(\pi-\theta)=\operatorname{cosec} \theta]
\begin{aligned} &\operatorname{cosec}\left(\pi-\frac{\pi}{4}\right)=\operatorname{cosec}\left(\frac{\pi}{4}\right) \\ &\operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2} \end{aligned}
By substituting these values in \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right) we get,
\operatorname{cosec}^{-1}(\sqrt{2})
Let, y=\operatorname{cosec}^{-1}(\sqrt{2})
\begin{aligned} &\operatorname{cosec} y=\sqrt{2} \\ &\operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2} \end{aligned}
The range of the principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text { and } \operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2}
\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{\pi}{4}\right)=\frac{\pi}{4}
Hence, \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right)=\frac{\pi}{4}

Inverse Trignometric Functions exercise 3.7 question 5 (iii)

Answer: -\frac{\pi }{5}
Hint: The range of principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Given: \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right)
Explanation:
First we solve \operatorname{cosec}\left(\frac{6 \pi}{5}\right)
\operatorname{cosec}\left(\frac{6 \pi}{5}\right)=\operatorname{cosec}\left(\pi+\frac{\pi}{5}\right)
As we know, \operatorname{cosec}(\pi+\theta)=-\operatorname{cosec} \theta
Then, \operatorname{cosec}\left(\pi+\frac{\pi}{5}\right)=-\operatorname{cosec}\left(\frac{\pi}{5}\right)
-\operatorname{cosec}\left(\frac{\pi}{5}\right)=\operatorname{cosec}\left(-\frac{\pi}{5}\right)
As \operatorname{cosec}(-x)=-\operatorname{cosec} x
By substituting this value in \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right) we get,
\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(-\frac{\pi}{5}\right)\right)
As we know, \operatorname{cosec}^{-1}(\operatorname{cosec} x)=x \text { when } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right)=-\frac{\pi}{5}

Inverse Trignometric Functions exercise 3.7 question 5 (iv)

Answer: -\frac{\pi }{6}
Hint: The range of principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Given: \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{11 \pi}{6}\right)
Explanation:
First we solve \operatorname{cosec}\left(\frac{11 \pi}{6}\right)
\operatorname{cosec}\left(\frac{11 \pi}{6}\right)=\operatorname{cosec}\left(2 \pi-\frac{\pi}{6}\right)
As we know, \operatorname{cosec}(2 \pi-\theta)=-\operatorname{cosec}(\theta)
\operatorname{cosec}\left(2 \pi-\frac{\pi}{6}\right)=-\operatorname{cosec}\left(\frac{\pi}{6}\right)
As we know, \operatorname{cosec}\left(\frac{\pi}{6}\right)=2
-\operatorname{cosec}\left(\frac{\pi}{6}\right)=-2
By substituting these values in \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{11 \pi}{6}\right) we get,
\operatorname{cosec}^{-1}(-2)
Let, y=\operatorname{cosec}^{-1}(-2)
\begin{aligned} &\operatorname{cosec} y=-2 \\ &-\operatorname{cosec} y=2 \\ &-\operatorname{cosec}\left(\frac{\pi}{6}\right)=2 \end{aligned}
As we know \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta
-\operatorname{cosec} \frac{\pi}{6}=\operatorname{cosec}\left(-\frac{\pi}{6}\right)
The range of principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text { and } \operatorname{cosec}\left(-\frac{\pi}{6}\right)=-2
\therefore \operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{11 \pi}{6}\right)\right) \mathrm{is}-\frac{\pi}{6}

Inverse Trignometric Functions exercise 3.7 question 5 (v)

Answer:

Answer: \frac{\pi }{6}
Hint: The range of principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Given: \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{13 \pi}{6}\right)
Explanation:
First we solve
\operatorname{cosec}\left(\frac{13 \pi}{6}\right)=\operatorname{cosec}\left(2 \pi+\frac{\pi}{6}\right)
As we know \operatorname{cosec}(2 \pi+\theta)=\operatorname{cosec} \theta
\csc (2\pi+\frac{\pi}{6})=\csc (\frac{\pi}{6})=2
By substituting these value in \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{13 \pi}{6}\right) we get,
\operatorname{cosec}^{-1}(2)
Now, \text { let } y=\operatorname{cosec}^{-1}(2)
\begin{aligned} &\operatorname{cosec} y=2 \\ &\operatorname{cosec}\left(\frac{\pi}{6}\right)=2 \end{aligned}
The range of the principal value in \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text { and } \operatorname{cosec}\left(\frac{\pi}{6}\right)=2
\begin{aligned} &\therefore \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{13 \pi}{6}\right) \text { is } \frac{\pi}{6} \\ &\therefore \operatorname{cosec}^{-1}(\operatorname{cosec} x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \end{aligned}

Question:5.6

Inverse Trignometric Functions exercise 3.7 question 5 (vi)

Answer: \frac{\pi }{4}
Hint: The range of principal value of \operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Given: \operatorname{cosec}^{-1}\left\{\operatorname{cosec}\left(-\frac{9 \pi}{4}\right)\right\}
Explanation:
First we solve \operatorname{cosec}\left(-\frac{9 \pi}{4}\right)
As we know \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta
\begin{aligned} &\operatorname{cosec}\left(-\frac{9 \pi}{4}\right)=-\operatorname{cosec}\left(\frac{9 \pi}{4}\right) \\ &-\operatorname{cosec}\left(\frac{9 \pi}{4}\right)=-\operatorname{cosec}\left(2 \pi+\frac{\pi}{4}\right) \\ &-\operatorname{cosec}\left(2 \pi+\frac{\pi}{4}\right)=-\operatorname{cosec}\left(\frac{\pi}{4}\right) \end{aligned}
As we know
\begin{aligned} &-\operatorname{cosec}(\theta)=\operatorname{cosec}(-\theta) \\ &-\operatorname{cosec}\left(\frac{\pi}{4}\right)=\operatorname{cosec}\left(-\frac{\pi}{4}\right) \end{aligned}
Now it becomes \operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{-\pi}{4}\right)
\begin{aligned} &\therefore \operatorname{cosec}^{-1}(\operatorname{cosec} x)=x \text { provide } \mathrm{d} x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \\ &\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{\pi}{4}\right)\right)=-\frac{\pi}{4} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (i)

Answer: \frac{\pi }{3}
Hint: The range of principal value of \cot ^{-1} is \left [ 0,\pi \right ]
Given: \cot ^{-1}\left(\cot \frac{\pi}{3}\right)
Explanation:
As we know
\cot ^{-1}(\cot x)=x \text { provided } \mathrm{} x \in[0, \pi]
By apply this situation we get,
\cot ^{-1}\left(\cot \frac{\pi}{3}\right)=\frac{\pi}{3}

Inverse Trignometric Functions exercise 3.7 question 6 (ii)

Answer: \frac{\pi }{3}
Hint: The range of principal value of \cot ^{-1}x is \left [ 0,\pi \right ]
Given: \cot ^{-1}\left(\cot \frac{4\pi}{3}\right)
Explanation:
First we solve \cot \frac{4 \pi}{3}
\cot \left(\frac{4 \pi}{3}\right)=\cot \left(\pi+\frac{\pi}{3}\right)
As we know \cot (\pi+\theta)=\cot \theta
\begin{aligned} &\cot \left(\pi+\frac{\pi}{3}\right)=\cot \frac{\pi}{3} \\ &\therefore \cot \frac{\pi}{3}=\frac{\sqrt{3}}{3} \end{aligned}
By substituting these value \cot ^{-1}\left(\cot \frac{4 \pi}{3}\right) we get,
\cot ^{-1}\left(\frac{\sqrt{3}}{3}\right)
Now, \text { let } y=\cot ^{-1}\left(\frac{\sqrt{3}}{3}\right)
\begin{aligned} &\cot y=\frac{\sqrt{3}}{3} \\ &\cot \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{3} \end{aligned}
The range of the principal value of \cot ^{-1} \text { is }[0, \pi]
\begin{aligned} &\cot ^{-1}\left(\cot \frac{4 \pi}{3}\right)=\frac{\pi}{3} \\ &\therefore \cot ^{-1}(\cot x)=x, x \in[0, \pi] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (iii)

Answer: \frac{\pi }{4}
Hint: The range of principal value of \cot ^{-1} is \left [ 0,\pi \right ]
Given: \cot ^{-1}\left(\cot \frac{9\pi}{4}\right)
Explanation:
First we solve \cot \frac{9 \pi}{4}
\begin{aligned} &\cot \frac{9 \pi}{4}=\cot \left(2 \pi+\frac{\pi}{4}\right) \\ &\therefore \cot (2 \pi+\theta)=\cot \theta \\ &\cot \left(2 \pi+\frac{\pi}{4}\right)=\cot \frac{\pi}{4} \\ &\therefore \cot \frac{\pi}{4}=1 \end{aligned}
By substituting these value in \cot ^{-1}\left(\cot \frac{9 \pi}{4}\right) we get,
\cot ^{-1}(1)
Now, \text { let } y=\cot ^{-1}(1)
\begin{aligned} &\cot y=1 \\ &\cot \left(\frac{\pi}{4}\right)=1 \end{aligned}
The range of principal value of \cot ^{-1} \text { is }[0, \pi] \text { and } \cot \left(\frac{\pi}{4}\right)=1
\begin{aligned} &\therefore \cot ^{-1}\left(\cot \frac{\pi}{4}\right)=\frac{\pi}{4} \\ &\cot ^{-1}(\cot x)=x, x \in[0, \pi] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (iv)

Answer: \frac{\pi }{6}
Hint: The range of principal value of \cot ^{-1} is \left [ 0,\pi \right ]
Given: \cot ^{-1}\left(\cot \frac{19\pi}{6}\right)
Explanation:
First we solve \cot \left(\frac{19 \pi}{6}\right)
\cot \left(\frac{19 \pi}{6}\right)=\cot \left(3 \pi+\frac{\pi}{6}\right)
As we know, \cot (n \pi+\theta)=\cot \theta
\begin{aligned} &\cot \left(3 \pi+\frac{\pi}{6}\right)=\cot \frac{\pi}{6} \\ &\cot \frac{\pi}{6}=\sqrt{3} \end{aligned} \left[\because \cot 30^{\circ}=\sqrt{3}\right]
By substituting these value in \cot ^{-1}\left(\cot \frac{19 \pi}{6}\right) we get,
\cot ^{-1}(\sqrt{3})
Now, \text { let } y=\cot ^{-1}(\sqrt{3})
\begin{aligned} &\cot y=\sqrt{3} \\ &\cot \left(\frac{\pi}{6}\right)=\sqrt{3} \end{aligned}
The range of the principal value of \cot ^{-1} \text { is }[0, \pi] \text { and } \cot \left(\frac{\pi}{6}\right)=\sqrt{3}
\therefore \cot ^{-1}\left(\cot \frac{\pi}{6}\right)=\frac{\pi}{6}
As we know, \cot ^{-1}(\cot x)=x, x \in[0, \pi]

Inverse Trignometric Functions exercise 3.7 question 6 (v)

Answer: \frac{\pi }{3}
Hint: The range of principal value of \cot ^{-1} is \left [ 0,\pi \right ]
Given: \cot ^{-1}\left\{\cot \left(-\frac{8 \pi}{3}\right)\right\}
Explanation:
First we solve \cot \left(-\frac{8 \pi}{3}\right)
As we know \cot (-\theta)=-\cot \theta
\begin{aligned} &\cot \left(-\frac{8 \pi}{3}\right)=-\cot \left(\frac{8 \pi}{3}\right) \\ &-\cot \left(\frac{8 \pi}{3}\right)=-\cot \left(3 \pi-\frac{\pi}{3}\right) \end{aligned}
As we know, \cot (2 \pi-\theta)=-\cot \theta
\begin{aligned} -\cot \left(3 \pi-\frac{\pi}{3}\right) &=-\cot \left(-\frac{\pi}{3}\right) \\ &=\cot \left(\frac{\pi}{3}\right) \\ \cot \left(\frac{\pi}{3}\right) &=\frac{1}{\sqrt{3}} \end{aligned}
By substituting these values in \cot ^{-1}\left(\cot \left(-\frac{8 \pi}{3}\right)\right) we get,
\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)
Now,
Let y=\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)
\begin{aligned} &\cot y=\frac{1}{\sqrt{3}} \\ &\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \end{aligned}
The range of the principal value of \cot ^{-1} \text { is }[0, \pi]
\begin{aligned} &\cot ^{-1}\left(\cot \frac{\pi}{3}\right)=\frac{\pi}{3} \\ &\therefore \cot ^{-1}(\cot x)=x, \text { where } x \in[0, \pi] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (vi)

Answer: \frac{\pi }{4}
Hint: The range of principal value of \cot ^{-1} is \left [ 0,\pi \right ]
Given: \cot ^{-1}\left\{\cot \left(\frac{21 \pi}{4}\right)\right\}
Explanation:
First we solve \cot \left(\frac{21 \pi}{4}\right)
\cot \left(\frac{21 \pi}{4}\right)=\cot \left(5 \pi+\frac{\pi}{4}\right)
As we know \cot (n \pi+\theta)=\cot \theta \text { where } n \text { is }(1,2,3, \ldots)
\cot \left(5 \pi+\frac{\pi}{4}\right)=\cot \left(\frac{\pi}{4}\right) \quad\left[\because \cot \frac{\pi}{4}=1\right]
\cot \left(\frac{\pi}{4}\right)=1
By substituting these values in \cot ^{-1}\left\{\cot \left(\frac{21 \pi}{4}\right)\right\} we get,
\cot ^{-1}(1)
Now , \text { let } y=\cot ^{-1}(1)
\begin{aligned} &\cot y=1 \\ &\cot \left(\frac{\pi}{4}\right)=1 \end{aligned}
The range of principal value of \cot ^{-1} \text { is }[0, \pi] \text { and } \cot \left(\frac{\pi}{4}\right)=1
\begin{aligned} &\therefore \cot ^{-1}\left(\cot \frac{\pi}{4}\right)=\frac{\pi}{4} \\ &\cot ^{-1}(\cot x)=x, x \in[0, \pi] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (i)

Answer: \sec ^{-1} \frac{x}{a}
Hint: The range of principal value of \cot ^{-1} \text { is }[0, \pi]
Given: \cot ^{-1}\left\{\frac{a}{\sqrt{x^{2}-a^{2}}}\right\},|x|>a
Explanation:
Let x=a \sec \theta
Now, we put value of x in given question.
\begin{aligned} &\cot ^{-1}\left\{\frac{a}{\sqrt{(a \sec \theta)^{2}-a^{2}}}\right\} \\ &\cot ^{-1}\left\{\frac{a}{\sqrt{a^{2} \sec ^{2} \theta-a^{2}}}\right\} \\ &\cot ^{-1}\left\{\frac{a}{\sqrt{a^{2}\left(\sec ^{2} \theta-1\right)}}\right\} \end{aligned}
As we know 1+\tan ^{2} \theta=\sec ^{2} \theta
\begin{aligned} &\sec ^{2} \theta-1=\tan ^{2} \theta \\ &\cot ^{-1}\left\{\frac{a}{\sqrt{a^{2} \tan ^{2} \theta}}\right\} \end{aligned}
Now we remove square root
\begin{aligned} &\cot ^{-1}\left\{\frac{a}{a \tan \theta}\right\} \\ &\cot ^{-1}\left\{\frac{1}{\tan \theta}\right\} \end{aligned}
As we know, \cot \theta=\frac{1}{\tan \theta}
\cot ^{-1}(\cot \theta)=\theta
As we know, \cot ^{-1}(\cot \theta), \theta \in[0, \pi]
\begin{aligned} &x=a \sec \theta \\ &\theta=\sec ^{-1} \frac{x}{a} \end{aligned}
Hence, \cot ^{-1}\left\{\frac{a}{\sqrt{x^{2}-a^{2}}}\right\},|x|>a \text { is } \sec ^{-1} \frac{x}{a}

Inverse Trignometric Functions exercise 3.7 question 7 (ii)

Answer: \frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \tan ^{-1}\left\{x+\sqrt{1+x^{2}}\right\}, x \in R
Explanation:
Let x=\cot \theta ; \theta=\cot ^{-1} x
Now ,
\tan ^{-1}\left\{x+\sqrt{1+x^{2}}\right\}=\tan ^{-1}\left\{\cot \theta+\sqrt{1+\cot ^{2} \theta}\right\}
As we know, 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta
\begin{aligned} &\tan ^{-1}\left\{\cot \theta+\sqrt{\operatorname{cosec}^{2} \theta}\right\} \\ &\tan ^{-1}\{\cot \theta+\operatorname{cosec} \theta\} \end{aligned}
As we know, \cot \theta=\frac{\cos \theta}{\sin \theta^{\prime}} \sin \theta=\frac{1}{\operatorname{cosec} \theta}
\begin{aligned} &\tan ^{-1}\left\{\frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}\right\} \\ &\tan ^{-1}\left\{\frac{\cos \theta+1}{\sin \theta}\right\} \end{aligned}
As we know, \sin 2 \theta=2 \sin \theta \cos \theta
\begin{aligned} &1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right) \\ &\tan ^{-1}\left\{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right\} \end{aligned}
\begin{aligned} &\tan ^{-1}\left\{\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right\} \\ &\tan ^{1}\left\{\cot \frac{\theta}{2}\right\} \end{aligned}
As we know \tan \theta=\cot \left(\frac{\pi}{2}-\theta\right)
\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}
As we know \tan ^{-1}(\tan x)=x \text { where } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\Rightarrow \quad\left(\frac{\pi}{2}-\frac{\theta}{2}\right)
Now put value of θ
\Rightarrow \quad \frac{\pi}{2}-\frac{\cot ^{-1} x}{2}

Inverse Trignometric Functions exercise 3.7 question 7 (iii)

Answer: \frac{1}{2} \cot ^{-1} x
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \tan ^{-1}\left\{\sqrt{1+x^{2}}-x\right\}, x \in R

Explanation:
Let x=\cot \theta ; \theta=\cot ^{-1} x
Now,
\tan ^{-1}\left\{\sqrt{1+x^{2}}-x\right\}=\tan ^{-1}\left\{\sqrt{1+\cot ^{2} \theta}-\cot \theta\right\}
As we know, 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta
\begin{aligned} &\tan ^{-1}\left\{\sqrt{\operatorname{cosec}^{2} \theta}-\cot \theta\right\} \\ &\tan ^{-1}\{\operatorname{cosec} \theta-\cot \theta\} \end{aligned}
As we know, \cot \theta=\frac{\cos \theta}{\sin \theta}, \operatorname{cosec} \theta=\frac{1}{\sin \theta}
\begin{aligned} &\tan ^{-1}\left\{\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right\} \\ &\tan ^{-1}\left\{\frac{1-\cos \theta}{\sin \theta}\right\} \end{aligned}
\\ \\ \hspace{0.5cm}\because 1-\cos x=2\sin^2\frac{x}{2}\\ \\ \hspace{0.5cm}\because \sin x=2sin \frac{x}{2} \cdot \cos \frac{x}{2}
\begin{aligned} &\tan ^{-1}\left\{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta }{2}}\right\} \\ &\tan ^{-1}\left[\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right] \\ &\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \end{aligned}
As we know, the principal range of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
\tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\begin{array}{ll} \Rightarrow & \frac{\theta}{2} \end{array}
\Rightarrow \; \; \; \; \frac{1}{2} \cot ^{-1} x

Inverse Trignometric Functions exercise 3.7 question 7 (iv)

Answer: \frac{1}{2} \tan ^{-1} x
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-1}{x}\right\}, x \neq 0
Explanation:
Let x=\tan \theta, \text { then } \theta=\tan ^{-1} x
\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)
As we know, 1+\tan ^{2} \theta=\sec ^{2} \theta
\begin{aligned} &\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right) \\ &\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \end{aligned}
\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right) \quad \because\left[\sec \theta=\frac{1}{\cos \theta}, \tan \theta=\frac{\sin \theta}{\cos \theta}\right]
\tan ^{-1}\left(\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}\right)
\tan ^{-1}\left(\frac{1-\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right)
\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)
\tan ^{-1}\left[\frac{2 \sin^{2} \frac{ \theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right] \quad \because\left[\begin{array}{c} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2} \end{array}\right]
\begin{aligned} &\tan ^{-1}\left[\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right] \\ &\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \end{aligned}
As we know \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\tan ^{-1}\left[\tan \frac{\theta}{2}\right]=\frac{\theta}{2}
Now putting the value of θ
\Rightarrow \quad \frac{1}{2} \tan ^{-1} x

Inverse Trignometric Functions exercise 3.7 question 7 (v)

Answer: \frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}+1}{x}\right\}, x \neq 0
Explanation:
Put x=\tan \theta, \text { then } \theta=\tan ^{-1} x
\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+1}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}+1}{\tan \theta}\right)
As we know, 1+\tan ^{2} \theta=\sec ^{2} \theta
\begin{aligned} &\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}+1}{\tan \theta}\right) \\ &\tan ^{-1}\left(\frac{\sec \theta+1}{\tan \theta}\right) \end{aligned}
\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}+1}{\frac{\sin \theta}{\cos \theta}}\right) \quad \because\left[\sec \theta=\frac{1}{\cos \theta}, \tan \theta=\frac{\sin \theta}{\cos \theta}\right]
\begin{aligned} &\tan ^{-1}\left(\frac{\frac{1+\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}\right) \\ &\tan ^{-1}\left(\frac{1+\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right) \\ &\tan ^{-1}\left(\frac{1+\cos \theta}{\sin \theta}\right) \end{aligned}
\tan ^{-1}\left[\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right] \quad \because\left[\begin{array}{c} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1+\cos \theta=2 \cos ^{2} \frac{\theta}{2} \end{array}\right]
\tan ^{-1}\left[\cot \frac{\theta}{2}\right]
\begin{aligned} &\therefore \tan \theta=\cot \left(\frac{\pi}{2}-\theta\right) \\ &\tan ^{-1}\left(\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \end{aligned}
As we know the range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]
\begin{aligned} & \tan ^{-1}(\tan x)=x, x \in\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\} \\ \Rightarrow & \frac{\pi}{2}-\frac{\theta}{2} \\ \Rightarrow & \frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (vi)

Answer: \frac{1}{2} \cos ^{-1} \frac{x}{a}
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \tan ^{-1} \sqrt{\frac{a-x}{a+x}},-a<x<a
Explanation:
Let x=a \cos \theta
then \theta=\cos ^{-1} \frac{x}{a}
Now,
\tan ^{-1} \sqrt{\frac{a-x}{a+x}}=\tan ^{-1} \sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}
\begin{aligned} &\tan ^{-1} \sqrt{\frac{a(1-\cos \theta)}{a(1+\cos \theta)}} \\ &\tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \end{aligned}
\tan ^{-1} \sqrt{\frac{2 \sin \frac{2}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \quad\left[\begin{array}{l} \left.\because \begin{array}{l} 1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right) \\ 1-\cos x=2 \sin ^{2}\left(\frac{x}{2}\right) \end{array}\right] \end{array}\right.
\tan ^{-1} \sqrt{\tan ^{2} \frac{\theta}{2}} \quad \because \frac{\sin \theta}{\cos \theta}=\tan \theta
\tan ^{-1}\left(\tan \frac{\theta}{2}\right)
As we know, \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\begin{aligned} &\Rightarrow \quad \frac{\theta}{2} \\ &\Rightarrow \quad \frac{1}{2} \cos ^{-1} \frac{x}{a} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (vii)

Answer: \frac{1}{2} \sin ^{-1} \frac{x}{a}
Hint: The range of principal value of \tan ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \tan ^{-1}\left[\frac{x}{a+\sqrt{a^{2}-x^{2}}}\right],-a<x<a
Explanation:
Put x=a \sin \theta, \text { then } \theta=\sin ^{-1} \frac{x}{a}
\tan ^{-1}\left[\frac{a \sin \theta}{a+\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right]
\tan ^{-1}\left[\frac{a \sin \theta}{a+\sqrt{a^{2}\left(1-\sin ^{2} \theta\right)}}\right] \quad \because\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]
\begin{aligned} &\tan ^{-1}\left[\frac{a \sin \theta}{a+\sqrt{a^{2} \cos ^{2} \theta}}\right] \\ &\tan ^{-1}\left[\frac{a \sin \theta}{a+a \cos \theta}\right] \end{aligned}
\begin{aligned} &\tan ^{-1}\left[\frac{a \sin \theta}{a(1+\cos \theta)}\right] \\ &\tan ^{-1}\left[\frac{\sin \theta}{1+\cos \theta}\right] \end{aligned}
\because \left[\begin{array}{l} \sin 2 \theta=2 \sin \theta \cos \theta \\ \cos 2 x=2 \cos ^{2} x-1 \end{array}\right]
\tan ^{-1}\left\{\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{1+2 \cos ^{2} \frac{\theta}{2}-1}\right\}
\begin{aligned} &\tan ^{-1}\left[\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right] \\ &\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \end{aligned}
As we know, \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\begin{aligned} &\Rightarrow \quad \frac{\theta}{2} \\ &\Rightarrow \quad \frac{1}{2} \sin ^{-1} \frac{x}{a} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (viii)

Answer: \frac{\pi}{4}+\sin ^{-1} x
Hint: The range of principal value of \sin ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \sin ^{-1}\left[\frac{x+\sqrt{1+x^{2}}}{\sqrt{2}}\right], \frac{1}{2}<x<\frac{1}{\sqrt{2}}
Explanation:
Let x=\sin \alpha
Then \alpha=\sin ^{-1} x
\begin{aligned} &\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}=\frac{\sin \alpha+\sqrt{1-\sin ^{2} \alpha}}{\sqrt{2}} \\ &\therefore \sin ^{2} \alpha+\cos ^{2} \alpha=1 \end{aligned}
Then, 1-\sin ^{2} \alpha=\cos ^{2} \alpha
\begin{aligned} &=\frac{\sin \alpha+\sqrt{\cos ^{2} \alpha}}{\sqrt{2}} \\ &=\frac{\sin \alpha+\sqrt{\cos ^{2} \alpha}}{\sqrt{2}} \end{aligned}
\begin{aligned} &=\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}+\cos \alpha \cdot \frac{1}{\sqrt{2}}\right) \\ &\therefore \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}, \sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \end{aligned}
\begin{aligned} &=\sin \alpha \cdot \cos \left(\frac{\pi}{4}\right)+\cos \alpha \cdot \sin \left(\frac{\pi}{4}\right) \\ &\therefore \sin A \cos B+\cos A \sin B=\sin (A+B) \end{aligned}
\sin \alpha \cos \left(\frac{\pi}{4}\right)+\cos \alpha \sin \left(\frac{\pi}{4}\right)=\sin \left(\alpha+\frac{\pi}{4}\right)
Then put value of \frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}=\sin \left(\alpha+\frac{\pi}{4}\right)
\sin ^{-1}\left(\sin \left(\alpha+\frac{\pi}{4}\right)\right)
As we know, \sin ^{-1}(\sin x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\begin{aligned} &=\alpha+\frac{\pi}{4} \\ &=\sin ^{-1} x+\frac{\pi}{4} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (ix)

Answer: \frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x
Hint: The range of principal value of \sin ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \sin ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right], 0<x<1
Explanation:
Let x=\cos \theta
Then
\theta=\cos ^{-1} x
Now,
\sin ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right]=\sin ^{-1}\left[\frac{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}{2}\right]
\therefore 1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right) \& 1-\cos x=2 \sin ^{2}\left(\frac{x}{2}\right)
\begin{aligned} &\sin ^{-1}\left\{\frac{\sqrt{2 \cos ^{2} \frac{\theta}{2}}+\sqrt{2 \sin ^{2} \frac{\theta}{2}}}{2}\right\} \\ &\sin ^{-1}\left\{\frac{\sqrt{2} \cos \frac{\theta}{2}+\sqrt{2} \sin \frac{\theta}{2}}{2}\right\} \end{aligned}
\begin{aligned} &\sin ^{-1}\left\{\frac{\sqrt{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}{2}\right\} \\ &\therefore\{\sqrt{2} \times \sqrt{2}=2\} \end{aligned}
\begin{aligned} &\sin ^{-1}\left\{\frac{\sqrt{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}{\sqrt{2} \times \sqrt{2}}\right\} \\ &\sin ^{-1}\left\{\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\sqrt{2}}\right\} \\ &\sin ^{-1}\left\{\frac{1}{\sqrt{2}} \sin \frac{\theta}{2}+\frac{1}{\sqrt{2}} \cos \frac{\theta}{2}\right\} \end{aligned}
\begin{aligned} &\sin ^{-1}\left\{\sin \left(\frac{\theta}{2}+\frac{\pi}{4}\right)\right\} \\ &\sin ^{-1}(\sin \theta)=\theta, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{aligned}
\begin{aligned} &=\frac{\theta}{2}+\frac{\pi}{2} \\ &=\frac{\cos ^{-1} x}{2}+\frac{\pi}{4} \\ &\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}=\frac{\cos ^{-1} x}{2}+\frac{\pi}{4} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (x

Answer: \sqrt{1-x^{2}}
Hint: The range of principal value of \sin ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given: \sin \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right]
Explanation:
First we solve 2 \tan ^{-} \sqrt{\frac{1-x}{1+x}}
Let x=\cos 2 A
Therefore,
2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}=2 \tan ^{-1} \sqrt{\frac{1-\cos 2 A}{1+\cos 2 A}}
\left[1-\cos 2 x=2 \sin ^{2} x \& 1+\cos 2 A=2 \cos ^{2} A\right]
=2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} A}{2 \cos ^{2} A}}
\begin{aligned} &=2 \tan ^{-1} \sqrt{\tan ^{2} A} \\ &=2 \tan ^{-1}(\tan A) \end{aligned}
As we know, \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
=2 \tan ^{-1}(\tan A)=2 A
Therefore,
\sin \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right]
=\sin 2 A \quad \therefore\left(1-\sin ^{2} \theta=\cos ^{2} \theta\right)
=\sqrt{1-\cos ^{2} 2 A}
Now put value of \cos 2 A=x
\sqrt{1-\cos ^{2} 2 A}=\sqrt{1-x^{2}}

In Chapter 3, Inverse Trigonometry for Class 12 Mathematics, about 14 exercises in total. In Exercise 3.7, there are many different functions that the students are asked to work out. The concepts include solving sine, cosine, tangent, cotangent, secant, and cosecant functions. There are fifty-six questions, including the subparts in exercise 3.7. And the RD Sharma Class 12 Solution Chapter 3 Exercise 3.7 book will make your work effortless.

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Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available
Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive. 

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available
Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction. 

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions. 

3 Jobs Available
Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Investment Director

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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