RD Sharma Class 12 Exercise 3.7 Inverse Trigonometric FunctionSolutions Maths

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

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.

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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: $\frac{π}{6}$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin \frac{π}{6})

Explanation:
We know that the value of

\sin \frac{π}{6}

\frac{1}{2}

By substituting this value in

\sin^{- 1} (\sin \frac{π}{6})

We get

\sin^{- 1} (\frac{1}{2})

Let,

\begin{aligned} y = \sin^{- 1} (\frac{1}{2}) \\ \sin y = \frac{1}{2} \\ \sin y = (\frac{1}{2}) \\ \sin (\frac{π}{6}) = \frac{1}{2} \end{aligned}

The range of principal value of

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

and

\sin (\frac{π}{6}) = \frac{1}{2}

Therefore,

\sin^{- 1} (\sin \frac{π}{6}) = \frac{π}{6}

Hence,

\begin{aligned} ∴ \sin^{- 1} (\sin θ) = θ with θ \in [- \frac{π}{2}, \frac{π}{2}] \\ \sin^{- 1} (\sin \frac{π}{6}) is \frac{π}{6} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (ii)

Answer: $- \frac{π}{6}$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin \frac{7 π}{6})

Explanation:
First, we solve

(\sin \frac{7 π}{6})

Sin \frac{7 π}{6} = \sin (π + \frac{π}{6})

As we know

\sin (\frac{π}{6}) = \frac{1}{2}

\sin (π + \frac{π}{6}) = \sin (- \frac{π}{6}) ∵ \sin [π + θ] = \sin (- θ)

\sin^{- 1} (\sin \frac{7 π}{6}) = \sin^{- 1} (- \frac{1}{2})

Now,

let y = \sin^{- 1} (- \frac{1}{2})

\begin{aligned} - \sin y = \frac{1}{2} \\ - \sin (\frac{π}{6}) = \frac{1}{2} \\ - \sin (\frac{π}{6}) = \sin (- \frac{π}{6}) \end{aligned}

The range of principal value of

\sin^{- 1} (- \frac{π}{2}, \frac{π}{2}) and \sin (- \frac{π}{6}) = - \frac{1}{2}

∴ \sin^{- 1} (\sin \frac{π}{6}) = \frac{π}{6} {As Sin \frac{7 π}{6} = \sin (π + \frac{π}{6})}

\begin{aligned} [\sin^{- 1} (\sin x)] = x \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (iii)

Answer: $\frac{π}{6}$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin \frac{5 π}{6})

Explanation:
First, we solve

\sin \frac{5 π}{6}

\frac{5 π}{6} = π - \frac{π}{6}

Then,

\begin{aligned} \sin (π - \frac{π}{6}) = \sin (\frac{π}{6}) \\ \sin (π - θ) = \sin (θ) \end{aligned}

We know that the value of

\sin \frac{π}{6}

\frac{1}{2}

\sin \frac{5 π}{6} = \frac{1}{2}

By substituting this value in

\sin^{- 1} (\sin \frac{5 π}{6})

we get

\sin^{- 1} (\frac{1}{2})

Let

y = \sin^{- 1} (\frac{1}{2})

\begin{aligned} \sin y = \frac{1}{2} \\ \sin (\frac{π}{6}) = \frac{1}{2} \end{aligned}

The range of principal value of

\sin^{- 1} is [- \frac{π}{2}, \frac{π}{2}] and \sin (\frac{π}{6}) = \frac{1}{2}

\begin{aligned} \sin^{- 1} (\sin \frac{π}{6}) = \frac{π}{6} \\ \sin^{- 1} (\sin \frac{5 π}{6}) = \frac{π}{6} ∵ [\sin^{- 1} (\sin θ) = θ] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (iv)

Answer:

Answer: $- \frac{π}{7}$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin \frac{13 π}{7})

Explanation:
First we solve

(\sin \frac{13 π}{7})

\frac{13 π}{7} = 2 π - \frac{π}{7}

\sin (2 π - \frac{π}{7}) = \sin (- \frac{π}{7}) ∵ [\sin (2 π - θ) = \sin (- θ)]

By substituting these value in

\sin^{- 1} (\sin \frac{13 π}{7}) = \sin^{- 1} (\sin - \frac{π}{7})

\sin^{- 1} (\sin x) = x with x \in [- \frac{π}{2}, \frac{π}{2}]

The range of principal value of

\sin^{- 1} is [- \frac{π}{2}, \frac{π}{2}]

\sin^{- 1} (\sin \frac{13 π}{7}) = - \frac{π}{7}

Inverse Trignometric Functions exercise 3.7 question 1 (v)

Answer: $- \frac{π}{8}$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:
Explanation:
First we solve

(\sin \frac{17 π}{8})

\begin{aligned} \sin \frac{17 π}{8} = \sin (2 π + \frac{π}{8}) \\ \sin (2 π + \frac{π}{8}) = \sin (\frac{π}{8}) ∵ [\sin (2 π + θ) = \sin (θ)] \end{aligned}

By substituting these values in

\sin^{- 1} (\sin \frac{17 π}{8})

we get

\sin^{- 1} (\sin \frac{π}{8})

\begin{matrix} As \sin^{- 1} (\sin x) = x with x \in [- \frac{π}{2}, \frac{π}{2}] \\ ∴ \sin^{- 1} (\sin \frac{17 π}{8}) = \frac{π}{8} \end{matrix}

Inverse Trignometric Functions exercise 3.7 question 1 (vi)

Answer: $- \frac{π}{8}$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} {(\sin - \frac{17 π}{8})}

Explanation:
First we solve

(\sin - \frac{17 π}{8})

As we know,

- \sin θ = \sin (- θ)

\begin{aligned} ∴ (\sin - \frac{17 π}{8}) = - \sin \frac{17 π}{8} \\ - \sin \frac{17 π}{8} = - \sin (2 π + \frac{π}{8}) \end{aligned}

- \sin (2 π + \frac{π}{8}) = - \sin (\frac{π}{8}) ∵ [\sin (2 π + θ) = \sin (θ)]

\begin{aligned} ∴ [- \sin (θ) = \sin (- θ)] \\ - \sin (\frac{π}{8}) = \sin (- \frac{π}{8}) \end{aligned}

By substituting these values in

\sin^{- 1} {(\sin - \frac{17 π}{8})}

we get,

\sin^{- 1} (\sin - \frac{π}{8})

\begin{matrix} As \sin^{- 1} (\sin x) = x with x \in [- \frac{π}{2}, \frac{π}{2}] \\ ∴ \sin^{- 1} (\sin - \frac{π}{8}) = - \frac{π}{8} \end{matrix}

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

$π - 3$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin 3)

Explanation:
We know that,

\sin^{- 1} (\sin x) = x with x \in [- \frac{π}{2}, \frac{π}{2}]

which is approximately equals to

[- 1.57, 1.57]

But here

x = 3

, which do not lie on the above range.
∴ We know that

\sin (π - x) = \sin (x)

\begin{aligned} \sin (π - 3) = \sin (3) also π - 3 \in [- \frac{π}{2}, \frac{π}{2}] \\ \sin^{- 1} (\sin 3) = \sin^{- 1} (\sin π - 3) \\ Then, \sin^{- 1} (\sin π - 3) = π - 3 \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (viii)

Answer:

$π - 4$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin 4)

Explanation:
We know that

\sin^{- 1} (\sin x) = x with x \in [- \frac{π}{2}, \frac{π}{2}]

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 (π - x) = \sin (x) \\ \sin (π - 4) = \sin (4) also π - 4 \in [- \frac{π}{2}, \frac{π}{2}] \\ \sin^{- 1} (\sin 4) = \sin^{- 1} (\sin π - 4) \end{aligned}

Then,

\sin^{- 1} (\sin π - 4) = π - 4

Inverse Trignometric Functions exercise 3.7 question 1 (ix)

Answer: $12 - 4 π$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin 12)

Explanation:
We know that

\sin^{- 1} (\sin x) = x with x \in [- \frac{π}{2}, \frac{π}{2}]

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 π - x) = \sin (- x)

Hence

\sin (2 n π - 12) = \sin (- 12)

Here,

\begin{aligned} n = 2 also 12 - 4 π \in [- \frac{π}{2}, \frac{π}{2}] \\ \sin^{- 1} (\sin 12) = 12 - 4 π \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 1 (x)

Answer: $π - 2$
Hint: The principal value branch of function

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given:

\sin^{- 1} (\sin 2)

Explanation:
We know that

\sin^{- 1} (\sin x) = x with x \in [- \frac{π}{2}, \frac{π}{2}]

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 (π - x) = \sin (x)

\begin{aligned} \sin (π - 2) = \sin (2) also π - 2 \in [- \frac{π}{2}, \frac{π}{2}] \\ \sin^{- 1} (\sin 2) = \sin^{- 1} (\sin π - 2) \\ ∴ \sin^{- 1} (\sin π - 2) = π - 2 \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 2 (i)

Answer: $\frac{π}{4}$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given:

\cos^{- 1} {\cos (- \frac{π}{4})}

Explanation:
We know that

\cos (- \frac{π}{4}) = \cos (\frac{π}{4}) ∵ [\cos (- θ) = \cos θ]

Also know that,

\cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}

By substituting these values in

\cos^{- 1} {\cos (- \frac{π}{4})}

we get,

\cos^{- 1} (\frac{1}{\sqrt{2}})

Let

y = \cos^{- 1} (\frac{1}{\sqrt{2}})

\cos y = \frac{1}{\sqrt{2}}

Hence, range of principal value of

\cos^{- 1} is [0, π] and \cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}

\begin{aligned} \cos^{- 1} {\cos (- \frac{π}{4})} = \frac{π}{4} \\ ∴ \cos^{- 1} (\cos x) = x, x \in [0, π] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 2 (ii)

Answer: $\frac{3 π}{4}$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given:

\cos^{- 1} (\cos \frac{5 π}{4})

Explanation:
First we solve

\cos \frac{5 π}{4}

\begin{aligned} \cos \frac{5 π}{4} = \cos (π + \frac{π}{4}) \\ \cos (π + \frac{π}{4}) = - \frac{1}{\sqrt{2}} ∵ [\cos (- θ) = \cos θ] \end{aligned}

As we know that

\cos \frac{π}{4} = \frac{1}{\sqrt{2}}

\cos (\frac{5 π}{4}) = - \frac{1}{\sqrt{2}}

By substituting these value in

\cos^{- 1} {\cos (\frac{5 π}{4})}

we get,

\cos^{- 1} (- \frac{1}{\sqrt{2}})

Since

[\cos (- θ) = \cos θ]

Let

y = \cos^{- 1} (- \frac{1}{\sqrt{2}})

\begin{aligned} \cos y = - \frac{1}{\sqrt{2}} \\ - \cos (\frac{π}{4}) = \frac{1}{\sqrt{2}} \\ \cos (π - \frac{π}{4}) = - \frac{1}{\sqrt{2}} \\ \cos (\frac{3 π}{4}) = - \frac{1}{\sqrt{2}} \end{aligned}

Hence, range of principal value of

\cos^{- 1} is [0, π] and \cos (\frac{3 π}{4}) = - \frac{1}{\sqrt{2}}

\cos^{- 1} {\cos (\frac{5 π}{4})} = \frac{3 π}{4} ∴ \cos^{- 1} (\cos x) = x, x \in [0, π]

Inverse Trignometric Functions exercise 3.7 question 2 (iii)

Answer: $\frac{2 π}{3}$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given: $\cos^{- 1} (\cos \frac{4 π}{3})$
Explanation:
First we solve

\cos (\frac{4 π}{3})

\begin{aligned} \cos \frac{4 π}{3} = \cos (π + \frac{π}{3}) \\ ∴ \cos [π + θ] = - \cos θ \end{aligned}

\begin{aligned} \cos (π + \frac{π}{3}) = - \frac{1}{2} \\ \cos (\frac{4 π}{3}) = - \frac{1}{2} \end{aligned}

By substituting these value in

\cos^{- 1} {\cos (\frac{4 π}{3})}

\cos^{- 1} (- \frac{1}{2})

Now,
Let

y = \cos^{- 1} (- \frac{1}{2})

\begin{aligned} \cos y = - \frac{1}{2} \\ - \cos (\frac{π}{3}) = \frac{1}{2} \\ \cos (π - \frac{π}{3}) = - \frac{1}{2} \\ \cos (\frac{2 π}{3}) = - \frac{1}{2} \end{aligned}

Hence, range of principal value of

\cos^{- 1} is [0, π] and \cos (\frac{2 π}{3}) = - \frac{1}{2}

\cos^{- 1} (\cos \frac{2 π}{3}) = \frac{2 π}{3}

Inverse Trignometric Functions exercise 3.7 question 2 (iv)

Answer: $\frac{π}{6}$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given: $\cos^{- 1} (\cos \frac{13 π}{6})$
Explanation:
First we solve

\cos \frac{13 π}{6}

\begin{aligned} \cos \frac{13 π}{6} = \cos (2 π + \frac{π}{6}) \\ ∴ [\cos (2 π + θ)] = \cos θ \\ \cos (2 π + \frac{π}{6}) = \cos (\frac{π}{6}) \end{aligned}

\begin{matrix} = \frac{\sqrt{3}}{2} \\ \cos (\frac{13 π}{6}) = \frac{\sqrt{3}}{2} \end{matrix}

By substituting these values in

\cos^{- 1} (\cos \frac{13 π}{6})

we get,

\cos^{- 1} (\frac{\sqrt{3}}{2})

Now,

let y = \cos^{- 1} (\frac{\sqrt{3}}{2})

\begin{aligned} \cos y = \frac{\sqrt{3}}{2} \\ \cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} \end{aligned}

Hence, range of principal value of

\cos^{- 1} is [0, π] and \cos \frac{π}{6} = \frac{\sqrt{3}}{2}

\begin{aligned} \cos^{- 1} (\cos \frac{π}{6}) = \frac{π}{6} \\ ∴ \cos^{- 1} (\cos x) = x, x \in [0, π] \\ \cos^{- 1} {\cos (\frac{13 π}{6})} = \frac{π}{6} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 2 (v)

Answer: 3
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given: $\cos^{- 1} (\cos 3)$
Explanation:
We know that

\cos^{- 1} (\cos 3) = 3, 3 \in [0, π] [\cos^{- 1} (\cos θ) = θ if 0 \leq θ \leq π]

Hence, the answer is

\cos^{- 1} (\cos 3) = 3

Inverse Trignometric Functions exercise 3.7 question 2 (vi)

Answer: $2 π - 4$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given: $\cos^{- 1} (\cos 4)$
Explanation:
We know that

\cos^{- 1} (\cos x) = x if x \in [0, π] \approx [0, 3.14]

And here $x = 4$ which do not lie in the above range

$\begin{aligned} \cos (2 π - x) = \cos (x) \\ \cos (2 π - 4) = \cos (4) so 2 π - 4 belongs in [0, π] \\ \cos^{- 1} (\cos 4) = 2 π - 4 \end{aligned}$

Inverse Trignometric Functions exercise 3.7 question 2 (vii)

Answer: $2 π - 5$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given: $\cos^{- 1} (\cos 5)$
Explanation:
We know that

\cos^{- 1} (\cos x) = x if x \in [0, π] \approx [0, 3.14]

And here

x = 5

which do not lie in the above range.
We know that

\cos (2 π - x) = \cos (x)

\cos (2 π - 5) = \cos (5) so 2 π - 5 belongs in [0, π]

Hence,

\cos^{- 1} (\cos 5) = 2 π - 5

Inverse Trignometric Functions exercise 3.7 question 2 (viii)

Answer: $4 π - 12$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given: $\cos^{- 1} (\cos 12)$
Explanation:
We know that

\cos^{- 1} (\cos x) = x if x \in [0, π] \approx [0, 3.14]

Here

x = 12

which do not lie in the above range.
We know that

\begin{aligned} \cos (2 n π - x) = \cos (x) \\ \cos (2 n π - 12) = \cos 12 \end{aligned}

Here,

n = 2

Also,

\begin{aligned} 4 π - 12 belongs in [0, π] \end{aligned}

\cos^{- 1} (\cos 12) = 4 π - 12

Inverse Trignometric Functions exercise 3.7 question 3 (i)

Answer: $\frac{π}{3}$
Hint: The principal value branch of function

\cos^{- 1}

[0, π]

Given: $\tan^{- 1} (\tan \frac{π}{3})$
Explanation:
As

\tan^{- 1} (\tan x) = x if x \in [- \frac{π}{2}, \frac{π}{2}]

By applying this situation
we get

\tan^{- 1} (\tan \frac{π}{3}) = \frac{π}{3}

Hence,

\tan^{- 1} (\tan \frac{π}{3}) is \frac{π}{3}

Inverse Trignometric Functions exercise 3.7 question 3 (ii)

Answer: $- \frac{π}{7}$
Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} (\tan \frac{6 π}{7})$
Explanation:

First we solve $\tan \frac{6 π}{7}$

$\begin{aligned} \tan \frac{6 π}{7} = \tan (π - \frac{π}{7}) \\ \tan (π - \frac{π}{7}) = - \tan \frac{π}{7} (\tan (π - θ) = - \tan θ) \end{aligned}$

We know

$\tan^{- 1} (\tan x) = x if x \in [- \frac{π}{2}, \frac{π}{2}]$

$\tan^{- 1} (\tan \frac{6 π}{7}) = - \frac{π}{7}$

Inverse Trignometric Functions exercise 3.7 question 3 (iii)

Answer: $\frac{π}{6}$
Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} (\tan \frac{7 π}{6})$
Explanation:
First we solve

\tan \frac{7 π}{6}

\tan \frac{7 π}{6} = \tan (π + \frac{π}{6})

As we know,

\tan (π + θ) = \tan θ

\begin{aligned} \tan (π + \frac{π}{6}) = \tan \frac{π}{6} \\ ∴ \tan \frac{π}{6} = \frac{1}{\sqrt{3}} \end{aligned}

By substituting this value in

\tan^{- 1} (\tan \frac{7 π}{6})

we get

\tan^{- 1} (\frac{1}{\sqrt{3}})

Now,

let \tan^{- 1} (\frac{1}{\sqrt{3}}) = y

\begin{aligned} \tan y = \frac{1}{\sqrt{3}} \\ \tan (\frac{π}{6}) = \frac{1}{\sqrt{3}} \end{aligned}

The range of principal value of

\tan^{- 1} is [- \frac{π}{2}, \frac{π}{2}] and \tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}

\tan^{- 1} (\tan \frac{7 π}{6}) = \frac{π}{6} As \tan^{- 1} (\tan x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]

Inverse Trignometric Functions exercise 3.7 question 3 (iv)

Answer: $\frac{π}{4}$
Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} (\tan \frac{9 π}{4})$
Explanation:
First we solve

\tan \frac{9 π}{4}

\tan \frac{9 π}{4} = \tan (2 π + \frac{π}{4})

∴ \tan (2 π + θ) = \tan θ

\begin{aligned} \tan (2 π + \frac{π}{4}) = \tan \frac{π}{4} \\ ∴ \tan \frac{π}{4} = 1 \\ \tan \frac{9 π}{4} = 1 \end{aligned}

By substituting this value in

\tan^{- 1} (\tan \frac{9 π}{4})

, we get

\tan^{- 1} (1)

let,

\tan^{- 1} (1) = y

\begin{aligned} \tan y = 1 \\ \tan (\frac{π}{4}) = 1 \end{aligned}

The range of principal value of

\tan^{- 1} is [- \frac{π}{2}, \frac{π}{2}] and \tan (\frac{π}{4}) = 1

\tan^{- 1} (\tan \frac{9 π}{4}) = \frac{π}{4} \tan^{- 1} (\tan x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]

Inverse Trignometric Functions exercise 3.7 question 3 (v)

Answer: 1
Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} (\tan 1)$
Explanation:
As we know

\tan^{- 1} (\tan x) = x if x \in [- \frac{π}{2}, \frac{π}{2}]

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 - π

Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} (\tan 2)$
Explanation:
As

\tan^{- 1} (\tan x) = x if x \in [- \frac{π}{2}, \frac{π}{2}]

But here

x = 2

which does not belong to above range

\begin{aligned} \tan (π - θ) = - \tan (θ) \\ \tan (θ - π) = \tan (θ) \\ \tan (2 - π) = \tan (2) \end{aligned}

Now,

2 - π

in the given range

[- \frac{π}{2}, \frac{π}{2}]

Hence, $\tan^{- 1} (\tan 2) = 2 - π$

Inverse Trignometric Functions exercise 3.7 question 3 (vii)

Answer:

4 - π

Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} (\tan 4)$
Explanation:
As

\tan^{- 1} (\tan x) = x if x \in [- \frac{π}{2}, \frac{π}{2}]

But here

x = 4

which does not belongs to above range

\begin{aligned} \tan (π - θ) = - \tan (θ) \\ \tan (θ - π) = \tan θ \\ \tan (4 - π) = \tan (4) \end{aligned}

Now,

4 - π \in [- \frac{π}{2}, \frac{π}{2}]

Hence,

\tan^{- 1} (\tan 4) = 4 - π

Inverse Trignometric Functions exercise 3.7 question 3 (viii)

Answer:

12 - 4 π

Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} (\tan 12)$
Explanation:
As

\tan^{- 1} (\tan x) = x if x \in [- \frac{π}{2}, \frac{π}{2}]

But here

x = 12

which does not belongs to above range

\begin{aligned} \tan (2 n π - θ) = - \tan (θ) \\ \tan (θ - 2 n π) = \tan θ \end{aligned}

Here,

n = 2

\tan (12 - 4 π) = \tan (12)

Now,

12 - 4 π \in [- \frac{π}{2}, \frac{π}{2}]

Hence,

Inverse Trignometric Functions exercise 3.7 question 4 (i)

Answer:

\frac{π}{3}

Hint: The range of principal value of

\sec^{- 1}

\forall x \in [0, π], x \neq \frac{π}{2}

Given: $\sec^{- 1} (\sec \frac{π}{3})$
Explanation:
As

\sec^{- 1} (\sec x) = x if x \in [0, π] - {\frac{π}{2}}

By applying this situation we get,

\sec^{- 1} (\sec \frac{π}{3}) = \frac{π}{3}

Hence,

\sec^{- 1} (\sec \frac{π}{3}) = \frac{π}{3}

Inverse Trignometric Functions exercise 3.7 question 4 (ii)

Answer:

\frac{2 π}{3}

Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} (\sec \frac{2 π}{3})$
Explanation:
As

\sec^{- 1} (\sec x) = x if x \in [0, π] - {\frac{π}{2}}

By applying this situation we get,

\sec^{- 1} (\sec \frac{2 π}{3}) = \frac{2 π}{3}

Hence,

\sec^{- 1} (\sec \frac{2 π}{3}) = \frac{2 π}{3}

Inverse Trignometric Functions exercise 3.7 question 4 (iii)

Answer:

\frac{3 π}{4}

Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} (\sec \frac{5 π}{4})$
Explanation:
First we solve

\sec \frac{5 π}{4}

But

(2 π - \frac{3 π}{4})

belongs to 3 quadrant
So,

\sec (2 π - \frac{3 π}{4}) = \sec \frac{3 π}{4}

\begin{aligned} ∴ \sec \frac{3 π}{4} = \sqrt{2} \\ - \sec \frac{3 π}{4} = - \sqrt{2} \end{aligned}

By substituting these value in

\sec^{- 1} (\sec \frac{5 π}{4})

we get,

\sec^{- 1} (- \sqrt{2})

Now,

let y = \sec^{- 1} (- \sqrt{2})

\Rightarrow \sec y = - \sqrt{2}

\Rightarrow - \sec (\frac{π}{4}) = \sqrt{2}

\begin{aligned} \Rightarrow \sec (π - \frac{π}{4}) \\ \Rightarrow \sec (\frac{3 π}{4}) = - \sqrt{2} \end{aligned}

The range of principal value of

\sec^{- 1} is [0, π] - {\frac{π}{2}}

\sec^{- 1} (\sec \frac{3 π}{4}) = \frac{3 π}{4} ∴ \sec^{- 1} (\sec x) = x, x \in [0, π] - {\frac{π}{2}}

Inverse Trignometric Functions exercise 3.7 question 4 (iv)

Answer:

\frac{π}{3}

Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} (\sec \frac{7 π}{3})$
Explanation:
First we solve

\sec \frac{7 π}{3}

\begin{aligned} \sec (\frac{7 π}{3}) = \sec (2 π + \frac{π}{3}) \\ ∴ [\sec (2 π + θ) = \sec θ] \end{aligned}

\begin{aligned} \sec (2 π + \frac{π}{3}) = \sec (\frac{π}{3}) \\ ∴ \sec (\frac{π}{3}) = 2 \end{aligned}

By substituting these value in

\sec^{- 1} (\sec \frac{7 π}{3})

we get,

\sec^{- 1} (2)

Now,

let y = \sec^{- 1} (2)

\begin{aligned} \sec y = 2 \\ \sec (\frac{π}{3}) = 2 \end{aligned}

The range of principal value of

\sec^{- 1} is [0, π] - {\frac{π}{2}} and \sec (\frac{π}{3}) = 2

\begin{aligned} \sec^{- 1} (\sec \frac{π}{3}) = \frac{π}{3} \\ \sec^{- 1} (\sec x) = x, x \in [0, π] - {\frac{π}{2}} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 4 (v)

Answer:

\frac{π}{5}

Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} (\sec \frac{9 π}{5})$
Explanation:
First we solve

\sec (\frac{9 π}{5})

\begin{aligned} \sec (\frac{9 π}{5}) = \sec (2 π - \frac{π}{5}) \\ ∴ [\sec (2 π - θ)] = \sec θ \end{aligned}

\begin{aligned} \sec (2 π - \frac{π}{5}) = \sec (\frac{π}{5}) \\ \sec (\frac{9 π}{5}) = \sec (\frac{π}{5}) \end{aligned}

By substituting these value in

\sec^{- 1} (\sec \frac{9 π}{5})

we get,

\begin{aligned} \sec^{- 1} (\sec \frac{π}{5}) \\ ∴ \sec^{- 1} (\sec x) = x, x \in [0, π] - {\frac{π}{2}} \\ \sec^{- 1} (\sec \frac{π}{5}) = \frac{π}{5} \end{aligned}

Hence,

\sec^{- 1} (\sec \frac{9 π}{5}) = \frac{π}{5}

Inverse Trignometric Functions exercise 3.7 question 4 (vi)

Answer: $\frac{π}{3}$
Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} {\sec (- \frac{7 π}{3})}$
Explanation:
First we solve

\sec (- \frac{7 π}{3})

As we know that

\sec (- θ) = \sec (θ)

\begin{aligned} \sec (- \frac{7 θ}{3}) = \sec (\frac{7 π}{3}) \\ \sec (\frac{7 π}{3}) = \sec (2 π + \frac{π}{3}) \end{aligned}

\begin{aligned} \sec (2 π + \frac{π}{3}) = \sec (\frac{π}{3}) ∵ [\sec (2 π + θ) = \sec θ] \\ \sec (\frac{π}{3}) = 2 \end{aligned}

By substituting these value in

\sec^{- 1} {\sec (- \frac{7 π}{3})}

we get,

\sec^{- 1} (2)

Now,

let y = \sec^{- 1} (2)

\begin{aligned} sec y = 2 \\ \sec (\frac{π}{3}) = 2 \end{aligned}

The range of principal value of

\sec^{- 1} is [0, π] - {\frac{π}{2}} and \sec (\frac{π}{3}) = 2

\begin{aligned} \sec^{- 1} (\sec \frac{π}{3}) = \frac{π}{3} \\ \sec^{- 1} (\sec x) = x, x \in [0, π] - {\frac{π}{2}} \end{aligned}

Hence,

\sec^{- 1} (\sec - \frac{7 π}{3}) = \frac{π}{3}

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

Answer:

Answer: $\frac{π}{3}$
Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} {\sec (- \frac{7 π}{3})}$

Inverse Trignometric Functions exercise 3.7 question 4 (vii)

Answer:

\frac{3 π}{4}

Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} (\sec \frac{13 π}{4})$
Explanation:
First we solve

\sec (\frac{13 π}{4})

\sec (\frac{13 π}{4}) = \sec (4 π - \frac{3 π}{4})

As we know

[\sec 2 π - θ] = \sec θ

But here after subtract

\frac{3 π}{4}

it comes in III quadrant where sec is negative

\begin{aligned} \sec (4 π - \frac{3 π}{4}) = - \sec (\frac{3 π}{4}) \\ ∴ \sec \frac{3 π}{4} = \sqrt{2} \\ - \sec (\frac{3 π}{4}) = - \sqrt{2} \end{aligned}

By substituting these values in

\sec^{- 1} (\sec \frac{13 π}{4})

we get,

\sec^{- 1} (- \sqrt{2})

Now,

let y = \sec^{- 1} (- \sqrt{2})

\begin{aligned} \sec y = - \sqrt{2} \\ - \sec y = \sqrt{2} \\ - \sec (\frac{π}{4}) = \sqrt{2} \\ \sec (π - \frac{π}{4}) = \sqrt{2} \\ \sec (\frac{3 π}{4}) = - \sqrt{2} \end{aligned}

The range of the principle value of

\sec^{- 1} is [0, π] - {\frac{π}{2}} and \sec (\frac{3 π}{4}) = - \sqrt{2}

\sec^{- 1} (\sec \frac{3 π}{4}) = \frac{3 π}{4}

∴ \sec^{- 1} (\sec x) = x, x \in [0, π] - {\frac{π}{2}}

Inverse Trignometric Functions exercise 3.7 question 4 (viii)

Answer: $\frac{π}{6}$
Hint: The range of principal value of

\sec^{- 1}

[0, π] - [\frac{π}{2}]

Given: $\sec^{- 1} (\sec \frac{25 π}{6})$
Explanation:
First we solve

\sec \frac{25 π}{6}

\sec (\frac{25 π}{6}) = \sec (4 π + \frac{π}{6})

As we know

\sec [2 π + θ] = \sec θ

\sec (4 π + \frac{π}{6}) = \sec (\frac{π}{6})

As we know

\sec (\frac{π}{6}) = \frac{2 \sqrt{3}}{3}

By substituting these value in

\sec^{- 1} (\sec \frac{25 π}{6})

we get,

\sec^{- 1} (\frac{2 \sqrt{3}}{3})

Now,

let y = \sec^{- 1} (\frac{2 \sqrt{3}}{3})

\begin{aligned} \sec y = \frac{2 \sqrt{3}}{3} \\ \sec (\frac{π}{6}) = \frac{2 \sqrt{3}}{3} \end{aligned}

The range of principal value of

\sec^{- 1} is [0, π] - {\frac{π}{2}} and \sec (\frac{π}{6}) = \frac{2 \sqrt{3}}{3}

\begin{aligned} \sec^{- 1} (\sec \frac{25 π}{6}) = \frac{π}{6} \\ \sec^{- 1} (\sec x) = x, x \in [0, π] - {\frac{π}{2}} \end{aligned}

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

Answer:

\frac{π}{4}

Hint: The range of principal value of

{cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0}

Given: ${cosec}^{- 1} (cosec \frac{π}{4})$
Explanation:
As we know,

{cosec}^{- 1} (cosec x) = x when x \in [- \frac{π}{2}, \frac{π}{2}] - {0}

By applying this situation we get,

{cosec}^{- 1} (cosec \frac{π}{4}) = \frac{π}{4}

Inverse Trignometric Functions exercise 3.7 question 5 (ii)

Answer: $\frac{π}{4}$
Hint: The range of principal value of ${cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0}$
Given: ${cosec}^{- 1} (cosec \frac{3 π}{4})$
Explanation:
First we solve

cosec (\frac{3 π}{4})

cosec (\frac{3 π}{4}) = cosec (π - \frac{π}{4})

As we know,

[cosec (π - θ) = cosec θ]

\begin{aligned} cosec (π - \frac{π}{4}) = cosec (\frac{π}{4}) \\ cosec (\frac{π}{4}) = \sqrt{2} \end{aligned}

By substituting these values in

{cosec}^{- 1} (cosec \frac{3 π}{4})

we get,

{cosec}^{- 1} (\sqrt{2})

Let,

y = {cosec}^{- 1} (\sqrt{2})

\begin{aligned} cosec y = \sqrt{2} \\ cosec (\frac{π}{4}) = \sqrt{2} \end{aligned}

The range of the principal value of

{cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0} and cosec (\frac{π}{4}) = \sqrt{2}

{cosec}^{- 1} (cosec \frac{π}{4}) = \frac{π}{4}

Hence,

{cosec}^{- 1} (cosec \frac{3 π}{4}) = \frac{π}{4}

Inverse Trignometric Functions exercise 3.7 question 5 (iii)

Answer: $- \frac{π}{5}$
Hint: The range of principal value of ${cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0}$
Given: ${cosec}^{- 1} (cosec \frac{6 π}{5})$
Explanation:
First we solve

cosec (\frac{6 π}{5})

cosec (\frac{6 π}{5}) = cosec (π + \frac{π}{5})

As we know,

cosec (π + θ) = - cosec θ

Then,

cosec (π + \frac{π}{5}) = - cosec (\frac{π}{5})

- cosec (\frac{π}{5}) = cosec (- \frac{π}{5})

cosec (- x) = - cosec x

By substituting this value in

{cosec}^{- 1} (cosec \frac{6 π}{5})

we get,

{cosec}^{- 1} (cosec (- \frac{π}{5}))

As we know,

{cosec}^{- 1} (cosec x) = x when x \in [- \frac{π}{2}, \frac{π}{2}] - {0}

{cosec}^{- 1} (cosec \frac{6 π}{5}) = - \frac{π}{5}

Inverse Trignometric Functions exercise 3.7 question 5 (iv)

Answer: $- \frac{π}{6}$
Hint: The range of principal value of ${cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0}$
Given: ${cosec}^{- 1} (cosec \frac{11 π}{6})$
Explanation:
First we solve

cosec (\frac{11 π}{6})

cosec (\frac{11 π}{6}) = cosec (2 π - \frac{π}{6})

As we know,

cosec (2 π - θ) = - cosec (θ)

cosec (2 π - \frac{π}{6}) = - cosec (\frac{π}{6})

As we know,

cosec (\frac{π}{6}) = 2

- cosec (\frac{π}{6}) = - 2

By substituting these values in

{cosec}^{- 1} (cosec \frac{11 π}{6})

we get,

{cosec}^{- 1} (- 2)

Let,

y = {cosec}^{- 1} (- 2)

\begin{aligned} cosec y = - 2 \\ - cosec y = 2 \\ - cosec (\frac{π}{6}) = 2 \end{aligned}

As we know

cosec (- θ) = - cosec θ

- cosec \frac{π}{6} = cosec (- \frac{π}{6})

The range of principal value of

{cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0} and cosec (- \frac{π}{6}) = - 2

∴ {cosec}^{- 1} (cosec (\frac{11 π}{6})) is - \frac{π}{6}

Inverse Trignometric Functions exercise 3.7 question 5 (v)

Answer:

Answer: $\frac{π}{6}$
Hint: The range of principal value of ${cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0}$
Given: ${cosec}^{- 1} (cosec \frac{13 π}{6})$
Explanation:
First we solve

cosec (\frac{13 π}{6}) = cosec (2 π + \frac{π}{6})

As we know

cosec (2 π + θ) = cosec θ

\csc (2 π + \frac{π}{6}) = \csc (\frac{π}{6}) = 2

By substituting these value in

{cosec}^{- 1} (cosec \frac{13 π}{6})

we get,

{cosec}^{- 1} (2)

Now,

let y = {cosec}^{- 1} (2)

\begin{aligned} cosec y = 2 \\ cosec (\frac{π}{6}) = 2 \end{aligned}

The range of the principal value in

{cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0} and cosec (\frac{π}{6}) = 2

\begin{aligned} ∴ {cosec}^{- 1} (cosec \frac{13 π}{6}) is \frac{π}{6} \\ ∴ {cosec}^{- 1} (cosec x) = x, x \in [- \frac{π}{2}, \frac{π}{2}] - {0} \end{aligned}

Question:5.6

Inverse Trignometric Functions exercise 3.7 question 5 (vi)

Answer: $\frac{π}{4}$
Hint: The range of principal value of ${cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0}$
Given: ${cosec}^{- 1} {cosec (- \frac{9 π}{4})}$
Explanation:
First we solve

cosec (- \frac{9 π}{4})

As we know

cosec (- θ) = - cosec θ

\begin{aligned} cosec (- \frac{9 π}{4}) = - cosec (\frac{9 π}{4}) \\ - cosec (\frac{9 π}{4}) = - cosec (2 π + \frac{π}{4}) \\ - cosec (2 π + \frac{π}{4}) = - cosec (\frac{π}{4}) \end{aligned}

As we know

\begin{aligned} - cosec (θ) = cosec (- θ) \\ - cosec (\frac{π}{4}) = cosec (- \frac{π}{4}) \end{aligned}

Now it becomes

{cosec}^{- 1} (cosec \frac{- π}{4})

\begin{aligned} ∴ {cosec}^{- 1} (cosec x) = x provide d x \in [- \frac{π}{2}, \frac{π}{2}] - {0} \\ {cosec}^{- 1} (cosec (\frac{π}{4})) = - \frac{π}{4} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (i)

Answer: $\frac{π}{3}$
Hint: The range of principal value of

\cot^{- 1}

[0, π]

Given: $\cot^{- 1} (\cot \frac{π}{3})$
Explanation:
As we know

\cot^{- 1} (\cot x) = x provided x \in [0, π]

By apply this situation we get,

\cot^{- 1} (\cot \frac{π}{3}) = \frac{π}{3}

Inverse Trignometric Functions exercise 3.7 question 6 (ii)

Answer: $\frac{π}{3}$
Hint: The range of principal value of

\cot^{- 1}

x

[0, π]

Given: $\cot^{- 1} (\cot \frac{4 π}{3})$
Explanation:
First we solve

\cot \frac{4 π}{3}

\cot (\frac{4 π}{3}) = \cot (π + \frac{π}{3})

As we know

\cot (π + θ) = \cot θ

\begin{aligned} \cot (π + \frac{π}{3}) = \cot \frac{π}{3} \\ ∴ \cot \frac{π}{3} = \frac{\sqrt{3}}{3} \end{aligned}

By substituting these value

\cot^{- 1} (\cot \frac{4 π}{3})

we get,

\cot^{- 1} (\frac{\sqrt{3}}{3})

Now,

let y = \cot^{- 1} (\frac{\sqrt{3}}{3})

\begin{aligned} \cot y = \frac{\sqrt{3}}{3} \\ \cot (\frac{π}{3}) = \frac{\sqrt{3}}{3} \end{aligned}

The range of the principal value of

\cot^{- 1} is [0, π]

\begin{aligned} \cot^{- 1} (\cot \frac{4 π}{3}) = \frac{π}{3} \\ ∴ \cot^{- 1} (\cot x) = x, x \in [0, π] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (iii)

Answer: $\frac{π}{4}$
Hint: The range of principal value of

\cot^{- 1}

[0, π]

Given: $\cot^{- 1} (\cot \frac{9 π}{4})$
Explanation:
First we solve

\cot \frac{9 π}{4}

\begin{aligned} \cot \frac{9 π}{4} = \cot (2 π + \frac{π}{4}) \\ ∴ \cot (2 π + θ) = \cot θ \\ \cot (2 π + \frac{π}{4}) = \cot \frac{π}{4} \\ ∴ \cot \frac{π}{4} = 1 \end{aligned}

By substituting these value in

\cot^{- 1} (\cot \frac{9 π}{4})

we get,

\cot^{- 1} (1)

Now,

let y = \cot^{- 1} (1)

\begin{aligned} \cot y = 1 \\ \cot (\frac{π}{4}) = 1 \end{aligned}

The range of principal value of

\cot^{- 1} is [0, π] and \cot (\frac{π}{4}) = 1

\begin{aligned} ∴ \cot^{- 1} (\cot \frac{π}{4}) = \frac{π}{4} \\ \cot^{- 1} (\cot x) = x, x \in [0, π] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (iv)

Answer: $\frac{π}{6}$
Hint: The range of principal value of

\cot^{- 1}

[0, π]

Given: $\cot^{- 1} (\cot \frac{19 π}{6})$
Explanation:
First we solve

\cot (\frac{19 π}{6})

\cot (\frac{19 π}{6}) = \cot (3 π + \frac{π}{6})

As we know,

\cot (n π + θ) = \cot θ

\begin{aligned} \cot (3 π + \frac{π}{6}) = \cot \frac{π}{6} \\ \cot \frac{π}{6} = \sqrt{3} \end{aligned}

[∵ \cot 30^{\circ} = \sqrt{3}]

By substituting these value in

\cot^{- 1} (\cot \frac{19 π}{6})

we get,

\cot^{- 1} (\sqrt{3})

Now,

let y = \cot^{- 1} (\sqrt{3})

\begin{aligned} \cot y = \sqrt{3} \\ \cot (\frac{π}{6}) = \sqrt{3} \end{aligned}

The range of the principal value of

\cot^{- 1} is [0, π] and \cot (\frac{π}{6}) = \sqrt{3}

∴ \cot^{- 1} (\cot \frac{π}{6}) = \frac{π}{6}

As we know,

\cot^{- 1} (\cot x) = x, x \in [0, π]

Inverse Trignometric Functions exercise 3.7 question 6 (v)

Answer: $\frac{π}{3}$
Hint: The range of principal value of

\cot^{- 1}

[0, π]

Given: $\cot^{- 1} {\cot (- \frac{8 π}{3})}$
Explanation:
First we solve

\cot (- \frac{8 π}{3})

As we know

\cot (- θ) = - \cot θ

\begin{aligned} \cot (- \frac{8 π}{3}) = - \cot (\frac{8 π}{3}) \\ - \cot (\frac{8 π}{3}) = - \cot (3 π - \frac{π}{3}) \end{aligned}

As we know,

\cot (2 π - θ) = - \cot θ

\begin{aligned} - \cot (3 π - \frac{π}{3}) & = - \cot (- \frac{π}{3}) \\ = \cot (\frac{π}{3}) \\ \cot (\frac{π}{3}) & = \frac{1}{\sqrt{3}} \end{aligned}

By substituting these values in

\cot^{- 1} (\cot (- \frac{8 π}{3}))

we get,

\cot^{- 1} (\frac{1}{\sqrt{3}})

Now,
Let

y = \cot^{- 1} (\frac{1}{\sqrt{3}})

\begin{aligned} \cot y = \frac{1}{\sqrt{3}} \\ \cot (\frac{π}{3}) = \frac{1}{\sqrt{3}} \end{aligned}

The range of the principal value of

\cot^{- 1} is [0, π]

\begin{aligned} \cot^{- 1} (\cot \frac{π}{3}) = \frac{π}{3} \\ ∴ \cot^{- 1} (\cot x) = x, where x \in [0, π] \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 6 (vi)

Answer: $\frac{π}{4}$
Hint: The range of principal value of

\cot^{- 1}

[0, π]

Given: $\cot^{- 1} {\cot (\frac{21 π}{4})}$
Explanation:
First we solve

\cot (\frac{21 π}{4})

\cot (\frac{21 π}{4}) = \cot (5 π + \frac{π}{4})

As we know

\cot (n π + θ) = \cot θ where n is (1, 2, 3, \dots)

\cot (5 π + \frac{π}{4}) = \cot (\frac{π}{4}) [∵ \cot \frac{π}{4} = 1]

\cot (\frac{π}{4}) = 1

By substituting these values in

\cot^{- 1} {\cot (\frac{21 π}{4})}

we get,

\cot^{- 1} (1)

Now ,

let y = \cot^{- 1} (1)

\begin{aligned} \cot y = 1 \\ \cot (\frac{π}{4}) = 1 \end{aligned}

The range of principal value of

\cot^{- 1} is [0, π] and \cot (\frac{π}{4}) = 1

\begin{aligned} ∴ \cot^{- 1} (\cot \frac{π}{4}) = \frac{π}{4} \\ \cot^{- 1} (\cot x) = x, x \in [0, π] \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} is [0, π]

Given: $\cot^{- 1} {\frac{a}{\sqrt{x^{2} - a^{2}}}}, | x | > a$
Explanation:
Let

x = a \sec θ

Now, we put value of

x

in given question.

\begin{aligned} \cot^{- 1} {\frac{a}{\sqrt{(a \sec θ)^{2} - a^{2}}}} \\ \cot^{- 1} {\frac{a}{\sqrt{a^{2} \sec^{2} θ - a^{2}}}} \\ \cot^{- 1} {\frac{a}{\sqrt{a^{2} (\sec^{2} θ - 1)}}} \end{aligned}

As we know

1 + \tan^{2} θ = \sec^{2} θ

\begin{aligned} \sec^{2} θ - 1 = \tan^{2} θ \\ \cot^{- 1} {\frac{a}{\sqrt{a^{2} \tan^{2} θ}}} \end{aligned}

Now we remove square root

\begin{aligned} \cot^{- 1} {\frac{a}{a \tan θ}} \\ \cot^{- 1} {\frac{1}{\tan θ}} \end{aligned}

As we know,

\cot θ = \frac{1}{\tan θ}

\cot^{- 1} (\cot θ) = θ

As we know,

\cot^{- 1} (\cot θ), θ \in [0, π]

\begin{aligned} x = a \sec θ \\ θ = \sec^{- 1} \frac{x}{a} \end{aligned}

Hence,

\cot^{- 1} {\frac{a}{\sqrt{x^{2} - a^{2}}}}, | x | > a is \sec^{- 1} \frac{x}{a}

Inverse Trignometric Functions exercise 3.7 question 7 (ii)

Answer: $\frac{π}{2} - \frac{1}{2} \cot^{- 1} x$
Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} {x + \sqrt{1 + x^{2}}}, x \in R$
Explanation:
Let

x = \cot θ; θ = \cot^{- 1} x

Now ,

\tan^{- 1} {x + \sqrt{1 + x^{2}}} = \tan^{- 1} {\cot θ + \sqrt{1 + \cot^{2} θ}}

As we know,

1 + \cot^{2} θ = {cosec}^{2} θ

\begin{aligned} \tan^{- 1} {\cot θ + \sqrt{{cosec}^{2} θ}} \\ \tan^{- 1} {\cot θ + cosec θ} \end{aligned}

As we know,

\cot θ = \frac{\cos θ}{\sin θ^{'}} \sin θ = \frac{1}{cosec θ}

\begin{aligned} \tan^{- 1} {\frac{\cos θ}{\sin θ} + \frac{1}{\sin θ}} \\ \tan^{- 1} {\frac{\cos θ + 1}{\sin θ}} \end{aligned}

∴ As we know,

\sin 2 θ = 2 \sin θ \cos θ

\begin{aligned} 1 + \cos x = 2 \cos^{2} (\frac{x}{2}) \\ \tan^{- 1} {\frac{2 \cos^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}} \end{aligned}

\begin{aligned} \tan^{- 1} {\frac{\cos \frac{θ}{2}}{\sin \frac{θ}{2}}} \\ \tan^{1} {\cot \frac{θ}{2}} \end{aligned}

As we know

\tan θ = \cot (\frac{π}{2} - θ)

\tan^{- 1} {\tan (\frac{π}{2} - \frac{θ}{2})}

As we know

\tan^{- 1} (\tan x) = x where x \in [- \frac{π}{2}, \frac{π}{2}]

\Rightarrow (\frac{π}{2} - \frac{θ}{2})

Now put value of θ

\Rightarrow \frac{π}{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}

[- \frac{π}{2}, \frac{π}{2}]

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

Explanation:
Let

x = \cot θ; θ = \cot^{- 1} x

Now,

\tan^{- 1} {\sqrt{1 + x^{2}} - x} = \tan^{- 1} {\sqrt{1 + \cot^{2} θ} - \cot θ}

As we know,

1 + \cot^{2} θ = {cosec}^{2} θ

\begin{aligned} \tan^{- 1} {\sqrt{{cosec}^{2} θ} - \cot θ} \\ \tan^{- 1} {cosec θ - \cot θ} \end{aligned}

As we know,

\cot θ = \frac{\cos θ}{\sin θ}, cosec θ = \frac{1}{\sin θ}

\begin{aligned} \tan^{- 1} {\frac{1}{\sin θ} - \frac{\cos θ}{\sin θ}} \\ \tan^{- 1} {\frac{1 - \cos θ}{\sin θ}} \end{aligned}

∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2} ∵ \sin x = 2 s i n \frac{x}{2} \cdot \cos \frac{x}{2}

\begin{aligned} \tan^{- 1} {\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}} \\ \tan^{- 1} [\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}] \\ \tan^{- 1} (\tan \frac{θ}{2}) \end{aligned}

As we know, the principal range of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

\tan^{- 1} (\tan x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]

\begin{array}{ll} \Rightarrow & \frac{θ}{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}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} {\frac{\sqrt{1 + x^{2}} - 1}{x}}, x \neq 0$
Explanation:
Let

x = \tan θ, then θ = \tan^{- 1} x

\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}) = \tan^{- 1} (\frac{\sqrt{1 + \tan^{2} θ} - 1}{\tan θ})

As we know,

1 + \tan^{2} θ = \sec^{2} θ

\begin{aligned} \tan^{- 1} (\frac{\sqrt{\sec^{2} θ} - 1}{\tan θ}) \\ \tan^{- 1} (\frac{\sec θ - 1}{\tan θ}) \end{aligned}

\tan^{- 1} (\frac{\frac{1}{\cos θ} - 1}{\frac{\sin θ}{\cos θ}}) ∵ [\sec θ = \frac{1}{\cos θ}, \tan θ = \frac{\sin θ}{\cos θ}]

\tan^{- 1} (\frac{\frac{1 - \cos θ}{\cos θ}}{\frac{\sin θ}{\cos θ}})

\tan^{- 1} (\frac{1 - \cos θ}{\cos θ} \times \frac{\cos θ}{\sin θ})

\tan^{- 1} (\frac{1 - \cos θ}{\sin θ})

\tan^{- 1} [\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}] ∵ [\begin{matrix} \sin 2 θ = 2 \sin θ \cos θ \\ 1 - \cos θ = 2 \sin^{2} \frac{θ}{2} \end{matrix}]

\begin{aligned} \tan^{- 1} [\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}] \\ \tan^{- 1} [\tan \frac{θ}{2}] \end{aligned}

As we know

\tan^{- 1} (\tan x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]

\tan^{- 1} [\tan \frac{θ}{2}] = \frac{θ}{2}

Now putting the value of θ

\Rightarrow \frac{1}{2} \tan^{- 1} x

Inverse Trignometric Functions exercise 3.7 question 7 (v)

Answer: $\frac{π}{2} - \frac{1}{2} \tan^{- 1} x$
Hint: The range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} {\frac{\sqrt{1 + x^{2}} + 1}{x}}, x \neq 0$
Explanation:
Put

x = \tan θ, then θ = \tan^{- 1} x

\tan^{- 1} (\frac{\sqrt{1 + x^{2}} + 1}{x}) = \tan^{- 1} (\frac{\sqrt{1 + \tan^{2} θ} + 1}{\tan θ})

As we know,

1 + \tan^{2} θ = \sec^{2} θ

\begin{aligned} \tan^{- 1} (\frac{\sqrt{\sec^{2} θ} + 1}{\tan θ}) \\ \tan^{- 1} (\frac{\sec θ + 1}{\tan θ}) \end{aligned}

\tan^{- 1} (\frac{\frac{1}{\cos θ} + 1}{\frac{\sin θ}{\cos θ}}) ∵ [\sec θ = \frac{1}{\cos θ}, \tan θ = \frac{\sin θ}{\cos θ}]

\begin{aligned} \tan^{- 1} (\frac{\frac{1 + \cos θ}{\cos θ}}{\frac{\sin θ}{\cos θ}}) \\ \tan^{- 1} (\frac{1 + \cos θ}{\cos θ} \times \frac{\cos θ}{\sin θ}) \\ \tan^{- 1} (\frac{1 + \cos θ}{\sin θ}) \end{aligned}

\tan^{- 1} [\frac{2 \cos^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}] ∵ [\begin{matrix} \sin 2 θ = 2 \sin θ \cos θ \\ 1 + \cos θ = 2 \cos^{2} \frac{θ}{2} \end{matrix}]

\tan^{- 1} [\cot \frac{θ}{2}]

\begin{aligned} ∴ \tan θ = \cot (\frac{π}{2} - θ) \\ \tan^{- 1} (\tan (\frac{π}{2} - \frac{θ}{2})) \end{aligned}

As we know the range of principal value of

\tan^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

\begin{aligned} \tan^{- 1} (\tan x) = x, x \in {- \frac{π}{2}, \frac{π}{2}} \\ \Rightarrow & \frac{π}{2} - \frac{θ}{2} \\ \Rightarrow & \frac{π}{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}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} \sqrt{\frac{a - x}{a + x}}, - a < x < a$
Explanation:
Let

x = a \cos θ

then

θ = \cos^{- 1} \frac{x}{a}

Now,

\tan^{- 1} \sqrt{\frac{a - x}{a + x}} = \tan^{- 1} \sqrt{\frac{a - a \cos θ}{a + a \cos θ}}

\begin{aligned} \tan^{- 1} \sqrt{\frac{a (1 - \cos θ)}{a (1 + \cos θ)}} \\ \tan^{- 1} \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} \end{aligned}

\tan^{- 1} \sqrt{\frac{2 \sin \frac{2}{2}}{2 \cos^{2} \frac{θ}{2}}} [\begin{array}{l} ∵ \begin{matrix} 1 + \cos x = 2 \cos^{2} (\frac{x}{2}) \\ 1 - \cos x = 2 \sin^{2} (\frac{x}{2}) \end{matrix}] \end{array}

\tan^{- 1} \sqrt{\tan^{2} \frac{θ}{2}} ∵ \frac{\sin θ}{\cos θ} = \tan θ

\tan^{- 1} (\tan \frac{θ}{2})

As we know,

\tan^{- 1} (\tan x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]

\begin{aligned} \Rightarrow \frac{θ}{2} \\ \Rightarrow \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}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\tan^{- 1} [\frac{x}{a + \sqrt{a^{2} - x^{2}}}], - a < x < a$
Explanation:
Put

x = a \sin θ, then θ = \sin^{- 1} \frac{x}{a}

\tan^{- 1} [\frac{a \sin θ}{a + \sqrt{a^{2} - a^{2} \sin^{2} θ}}]

\tan^{- 1} [\frac{a \sin θ}{a + \sqrt{a^{2} (1 - \sin^{2} θ)}}] ∵ [\sin^{2} θ + \cos^{2} θ = 1]

\begin{aligned} \tan^{- 1} [\frac{a \sin θ}{a + \sqrt{a^{2} \cos^{2} θ}}] \\ \tan^{- 1} [\frac{a \sin θ}{a + a \cos θ}] \end{aligned}

\begin{aligned} \tan^{- 1} [\frac{a \sin θ}{a (1 + \cos θ)}] \\ \tan^{- 1} [\frac{\sin θ}{1 + \cos θ}] \end{aligned}

∵ [\begin{array}{l} \sin 2 θ = 2 \sin θ \cos θ \\ \cos 2 x = 2 \cos^{2} x - 1 \end{array}]

\tan^{- 1} {\frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{1 + 2 \cos^{2} \frac{θ}{2} - 1}}

\begin{aligned} \tan^{- 1} [\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}] \\ \tan^{- 1} [\tan \frac{θ}{2}] \end{aligned}

As we know,

\tan^{- 1} (\tan x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]

\begin{aligned} \Rightarrow \frac{θ}{2} \\ \Rightarrow \frac{1}{2} \sin^{- 1} \frac{x}{a} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (viii)

Answer: $\frac{π}{4} + \sin^{- 1} x$
Hint: The range of principal value of

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\sin^{- 1} [\frac{x + \sqrt{1 + x^{2}}}{\sqrt{2}}], \frac{1}{2} < x < \frac{1}{\sqrt{2}}$
Explanation:
Let

x = \sin α

Then

α = \sin^{- 1} x

\begin{aligned} \frac{x + \sqrt{1 - x^{2}}}{\sqrt{2}} = \frac{\sin α + \sqrt{1 - \sin^{2} α}}{\sqrt{2}} \\ ∴ \sin^{2} α + \cos^{2} α = 1 \end{aligned}

Then,

1 - \sin^{2} α = \cos^{2} α

\begin{aligned} = \frac{\sin α + \sqrt{\cos^{2} α}}{\sqrt{2}} \\ = \frac{\sin α + \sqrt{\cos^{2} α}}{\sqrt{2}} \end{aligned}

\begin{aligned} = (\sin α \cdot \frac{1}{\sqrt{2}} + \cos α \cdot \frac{1}{\sqrt{2}}) \\ ∴ \cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}, \sin (\frac{π}{4}) = \frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} = \sin α \cdot \cos (\frac{π}{4}) + \cos α \cdot \sin (\frac{π}{4}) \\ ∴ \sin A \cos B + \cos A \sin B = \sin (A + B) \end{aligned}

\sin α \cos (\frac{π}{4}) + \cos α \sin (\frac{π}{4}) = \sin (α + \frac{π}{4})

Then put value of

\frac{x + \sqrt{1 - x^{2}}}{\sqrt{2}} = \sin (α + \frac{π}{4})

\sin^{- 1} (\sin (α + \frac{π}{4}))

As we know,

\sin^{- 1} (\sin x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]

\begin{aligned} = α + \frac{π}{4} \\ = \sin^{- 1} x + \frac{π}{4} \end{aligned}

Inverse Trignometric Functions exercise 3.7 question 7 (ix)

Answer: $\frac{π}{4} + \frac{1}{2} \cos^{- 1} x$
Hint: The range of principal value of

\sin^{- 1}

[- \frac{π}{2}, \frac{π}{2}]

Given: $\sin^{- 1} [\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}], 0 < x < 1$
Explanation:
Let

x = \cos θ

Then

θ = \cos^{- 1} x

Now,

\sin^{- 1} [\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}] = \sin^{- 1} [\frac{\sqrt{1 + \cos θ} + \sqrt{1 - \cos θ}}{2}]

∴ 1 + \cos x = 2 \cos^{2} (\frac{x}{2}) & 1 - \cos x = 2 \sin^{2} (\frac{x}{2})

\begin{aligned} \sin^{- 1} {\frac{\sqrt{2 \cos^{2} \frac{θ}{2}} + \sqrt{2 \sin^{2} \frac{θ}{2}}}{2}} \\ \sin^{- 1} {\frac{\sqrt{2} \cos \frac{θ}{2} + \sqrt{2} \sin \frac{θ}{2}}{2}} \end{aligned}

\begin{aligned} \sin^{- 1} {\frac{\sqrt{2} (\cos \frac{θ}{2} + \sin \frac{θ}{2})}{2}} \\ ∴ {\sqrt{2} \times \sqrt{2} = 2} \end{aligned}

\begin{aligned} \sin^{- 1} {\frac{\sqrt{2} (\cos \frac{θ}{2} + \sin \frac{θ}{2})}{\sqrt{2} \times \sqrt{2}}} \\ \sin^{- 1} {\frac{\cos \frac{θ}{2} + \sin \frac{θ}{2}}{\sqrt{2}}} \\ \sin^{- 1} {\frac{1}{\sqrt{2}} \sin \frac{θ}{2} + \frac{1}{\sqrt{2}} \cos \frac{θ}{2}} \end{aligned}

\begin{aligned} \sin^{- 1} {\sin (\frac{θ}{2} + \frac{π}{4})} \\ \sin^{- 1} (\sin θ) = θ, θ \in [- \frac{π}{2}, \frac{π}{2}] \end{aligned}

\begin{aligned} = \frac{θ}{2} + \frac{π}{2} \\ = \frac{\cos^{- 1} x}{2} + \frac{π}{4} \\ \sin^{- 1} {\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}} = \frac{\cos^{- 1} x}{2} + \frac{π}{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 $[- \frac{π}{2}, \frac{π}{2}]$

Given: $\sin [2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}]$
Explanation:
First we solve $2 \tan^{-} \sqrt{\frac{1 - x}{1 + x}}$
Let $x = \cos 2 A$
Therefore,
$2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} = 2 \tan^{- 1} \sqrt{\frac{1 - \cos 2 A}{1 + \cos 2 A}}$
$[1 - \cos 2 x = 2 \sin^{2} x & 1 + \cos 2 A = 2 \cos^{2} A]$
$= 2 \tan^{- 1} \sqrt{\frac{2 \sin^{2} A}{2 \cos^{2} A}}$
$\begin{aligned} = 2 \tan^{- 1} \sqrt{\tan^{2} A} \\ = 2 \tan^{- 1} (\tan A) \end{aligned}$
As we know, $\tan^{- 1} (\tan x) = x, x \in [- \frac{π}{2}, \frac{π}{2}]$
$= 2 \tan^{- 1} (\tan A) = 2 A$
Therefore,
$\sin [2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}]$
$= \sin 2 A ∴ (1 - \sin^{2} θ = \cos^{2} θ)$
$= \sqrt{1 - \cos^{2} 2 A}$
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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