#### Explain solution RD Sharma class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 3 sub question (i) maths

$\frac{-2\pi }{3}$

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$\frac{-2\pi }{3}$