RD Sharma Class 12 Exercise 3.5 Inverse trigonometric functions Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 3.5 Inverse trigonometric functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.5 Inverse trigonometric functions Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 21 Jan 2022, 12:18 PM IST

The RD Sharma Class 12 Solutions are increasingly used by many 12th graders to prepare for their public exams. The concepts in class 12 mathematics are hard for the students to perform. Reference materials like the RD Sharma Class 12 Solutions Chapter 3 Exercise 3.5 play a major role in making them understand the concepts in a better way.

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

Inverse trigonometric functions Ex 3.5

Inverse Trigonometric Functions exercise 3.5 question 1(i)

Answer:$\frac{-\pi}{4}$
Hints: The $\operatorname{cosec}^{-1}$ function is defined as a function whose domain is R-(-1,1) and the principal value branch of the function $\operatorname{cosec}^{-1}$ is -$\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Given: $\operatorname{cosec}^{-1}(-\sqrt{2})$
Explanation: Let $y=\operatorname{cosec}^{-1}(-\sqrt{2})$
$\therefore y=-\operatorname{cosec}^{-1}(\sqrt{2})$ $\left[\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}(x)\right]$
$\operatorname{cosec} y=-(\sqrt{2})$
$\operatorname{cosec} y =- \operatorname{cosec}\frac{\pi}{4}$ $\left[\because \operatorname{cosec} \frac{\pi}{4}=\sqrt{2}\right]$
$y = \frac{-\pi }{4}$
since Range of $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Therefore, principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ is -$\frac{-\pi}{4}$

Inverse Trigonometric Functions exercise 3.5 question 1(ii)

Answer:

$\frac{-\pi}{6}$
Hints: Domain of $\operatorname{cosec}^{-1}$ s R- (-1,1) and the range of $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Given: $\operatorname{cosec}^{-1}(-\sqrt{2})$
Explanation: Let $y=\operatorname{cosec}^{-1}(-\sqrt{2})$
$\operatorname{cosec} y=-2$
$\operatorname{cosec} y =- \operatorname{cosec}\frac{\pi}{6}$ $\left [\operatorname{cosec}\frac{-\pi}{6}=2 \right ]$
$\therefore y =\frac{-\pi}{6}$
Since, range of $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Therefore, principal value is $\frac{-\pi}{6}$

Inverse Trigonometric Functions exercise 3.5 question 1(iii)

Answer:

$\frac{-\pi}{3}$
Hints: The $\operatorname{cosec}^{-1}$ function is a function whose domain is R-(-1,1) and The principle value branch of $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Given: $\operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )$
Explanation: let $y=\operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )$
$\operatorname{cosec} y = \left ( \frac{2}{\sqrt{3}} \right )$
$\operatorname{cosec} y = \operatorname{cosec} \left ( \frac{\pi}{\sqrt{3}} \right )$ $\left[\because \operatorname{cosec} \frac{\pi}{3}=\frac{2}{\sqrt{3}}\right]$
$\therefore y = \frac{\pi }{3}$
Hence the principal value of $\operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )$ is $\frac{\pi}{3}$

Inverse Trigonometric Functions exercise 3.5 question 1 (iv)

Answer:

$\frac{-\pi}{2}$
Hints: $\operatorname{cosec}^{-1}$ is a function whose range of principal branch is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Thus, $\operatorname{cosec}^{-1}$ : R-(-1,1)→$\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Given: $\operatorname{cosec}^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$
Explanation: let $y = \left(2 \cos \frac{2 \pi}{3}\right)$
$\operatorname{cosec}y = \left(2 \cos \frac{2 \pi}{3}\right)$
Range of function $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Thus, $\operatorname{cosec}y = \left(2 \cos \frac{2 \pi}{3}\right)$
$\operatorname{cosec}y = 2 \times \frac{-1}{2}$ $\left [\cos \frac{2\pi }{3}= \frac{-1}{2} \right ]$
$\operatorname{cosec}y =-1$
$\begin{aligned} &\operatorname{cosec} y=\operatorname{cosec}\left(\frac{-\pi}{2}\right)\\ &y=\frac{-\pi}{2} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right], y \neq 0\\ &\text { Hence principal value of } \operatorname{cosec}^{-1}\left(2 \cos \frac{2 \pi}{3}\right) \text { is } \frac{-\pi}{2} \end{aligned}$

Inverse Trigonometric Functions exercise 3.5 question 2

Answer:

$\phi$
Hints: The $\operatorname{cosec}^{-1}$ function is defined as a function whose domain is R-(-1,1) and The principal value branch of the function $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Given: $\operatorname{cosec}^{-1}\left ( \frac{\sqrt{3}}{2} \right )$
Explanation: Domain of $\operatorname{cosec}^{-1} \mathrm{x} \text { is }[-\infty, 1] U[1, \infty]$
Let,$y=\operatorname{cosec}^{-1}\left ( \frac{\sqrt{3}}{2} \right )$
But $\frac{\sqrt{3}}{2} <1$
Therefore, it cannot be a value of y
Hence, set value of $\operatorname{cosec}^{-1}\left ( \frac{\sqrt{3}}{2} \right )$ is a null set $\phi$

Inverse Trigonometric Functions exercise 3.5 question 3(i)

Answer:

$\frac{-2\pi }{3}$
Hints: The principal value branch of the function $\sin ^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
The principal value branch of the function $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Then find all the principal values b/w these intervals.
Given:$\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)$
Explanation: $\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ …(i)
Let us solve for $\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
Let $x=\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
$\frac{-\sqrt{3}}{2}$
$\sin x = \sin \frac{\pi }{3}$ $\left [\because \sin \frac{\pi}{3}= \frac{\sqrt{3}}{2} \right ]$
$x = -\frac{\pi }{3}$
$\therefore \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ … (ii)
Let us solve for $\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)$
Let $y=\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)$
$\operatorname{cosec} y =\left(\frac{-2}{\sqrt{3}}\right)$
$\operatorname{cosec} y = \operatorname{cosec} \frac{\pi}{3}$ $\left [\because \operatorname{cosec} \frac{\pi}{3}= \frac{2}{\sqrt{3}} \right ]$
$y= \frac{-\pi}{3}$
$\therefore \operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3} \in\left[\frac{-\pi}{3}, \frac{\pi}{3}\right]$ ….(iii)
Now from equation (i)
$\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)$
Putting the values of $\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\text{ and }\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ [from (ii) (iii)]
Hence$\left ( \frac{-\pi }{3} \right )+\left ( \frac{-\pi}{3} \right )$
$\frac{-\pi }{3} - \frac{-\pi}{3}$
$\frac{-2\pi }{3}$
Therefore, the principal value of
$\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ is$\frac{-2\pi }{3}$

Inverse Trigonometric Functions exercise 3.5 question 3 (ii)

Answer:

$\frac{-\pi}{4}$
The principal value branch of the function $\sec ^{-1} \text { is }[0, \pi]-\left[\frac{-\pi}{2}\right]$
The principal value branch of the function $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Then find all the principal values between these intervals
Given:$\sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})$
Explanation: $\sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})$ ….(i)
Let us first solve for $\sec ^{-1}(\sqrt{2})$
Let $x=\sec ^{-1}(\sqrt{2})$
$\sec x= (\sqrt{2})$
$\sec x=\sec \frac{\pi}{4}$ $\left [\because \sec \frac{\pi}{4}= \sqrt{2} \right ]$
$x = \frac{\pi}{4}$
$\therefore \sec ^{-1}(\sqrt{2})=\frac{\pi}{4} \in[0, \pi]-\left[\frac{-\pi}{2}\right]$
The principal value of $\sec ^{-1}(\sqrt{2})$ is $\frac{\pi}{4}$ …. (ii)
Now let us find $2 \operatorname{cosec}^{-1}(-\sqrt{2})$
Let $y= 2 \operatorname{cosec}^{-1}(-\sqrt{2})$
$\operatorname{cosec} y= 2 x (\sqrt{2})$
$\operatorname{cosec} y= 2 \operatorname{cosec} \left ( \frac{\pi}{4} \right )$
$y = 2\times\frac{-\pi }{4}$
$y = \frac{-\pi }{2}$
$\therefore 2 \operatorname{cosec}^{-1}(-\sqrt{2})=\frac{-\pi}{2} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
The principal value of $2 \operatorname{cosec}^{-1}(-\sqrt{2})$ is-$\frac{-\pi}{2}$ …. (iii)
Now from equation (i)
$\sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})$
Putting the values of $\sec ^{-1}(\sqrt{2}) \: \text{and }2 \operatorname{cosec}^{-1}(-\sqrt{2})$ [from (ii) and (iii)]
$\therefore \frac{\pi}{4}+\left ( \frac{-\pi }{2} \right )$
$\frac{\pi}{4}- \frac{\pi }{2}$
$\frac{\pi-2\pi}{4}$
$\frac{-\pi}{4}$
Therefore, the principal value of $\sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})$ is $\frac{-\pi}{4}$

Inverse Trigonometric Functions exercise 3.5 question 3(iii)

Answer:

$\frac{\pi}{6}$
Hints: The principal value branch of the function $\operatorname{cosec}^{-1}$is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
The principal value branch of the function $\sin ^{-1}$is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]$
Given: $\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right]$
Range of $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
$\sin ^{-1}\left[\cos \left\{2 \times \frac{-\pi}{6}\right\} .\right.$ $\operatorname{cosec}^{-1}(\sqrt{2})= \frac{-\pi}{6}$
$\sin ^{-1}\left[\cos \left\{\frac{-\pi}{3}\right\}\right]$
$\sin ^{-1}\left[\cos \left\{\frac{-\pi}{3}\right\}\right]$ $[\cos( -\theta) = \cos \theta]$
$\begin{aligned} &\sin ^{-1}\left(\frac{1}{2}\right)\\ &\text { Let } y=\sin ^{-1}\left(\frac{1}{2}\right)\\ &\sin y=\sin \frac{\pi}{6}\\ &y=\frac{\pi}{6}\\ &\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\\ &\text { Therefore, the principal value of }\\ &\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right] \text { is } \frac{\pi}{6} \end{aligned}$

Inverse Trigonometric Functions exercise 3.5 question 3(iv)

Answer:

$\frac{-\pi}{3}$
Hints: The principal value branch of the function $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$
Given: $\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$
Explanation:
$\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$
$\operatorname{cosec}^{-1}\left ( 2 \times \left ( \frac{-1}{\sqrt{3}} \right ) \right )$ $\left [\because \tan \frac{11 \pi}{6}= \frac{-1}{\sqrt{3}} \right ]$
$\begin{aligned} &\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right) \\ &\text { Let, } y=\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right) \\ &\qquad \operatorname{cosec} y=\left(\frac{-2}{\sqrt{3}}\right) \end{aligned}$
$\operatorname{cosec} y = \operatorname{cosec} \frac{-\pi}{3}$ $\left [ \operatorname{cosec} \frac{\pi}{3}=\frac{2}{\sqrt{3}} \right ]$
$\therefore y= \frac{-\pi}{3}$
$\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3} \quad \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
Therefore, the principal value of $\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$ is $\frac{-\pi}{3}$

The third chapter in the class 12 mathematics book consists of 14 exercises. The fifth exercise of this chapter, ex 3.5 in particular consists of 9 questions given in the textbook. The topics covered in the RD Sharma class 12th exercise 3.5 'Inverse Trigonometry' portion include concepts like inverse of cosine, tangent, secant, sine, cosecant and cotangent functions. All the solved sums on these concepts can be found at the RD Sharma Solutions Class 12 Chapter 3 Exercise 3.5 book.

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