#### Provide solution for RD Sharma maths Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 2 Sub question (ii) maths textbook solution.

Answer : $\frac{\pi }{3}$

Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.

Thus, $\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

The branch with range $\left [ 0,\pi \right ]$ is called the principal value branch of the function $\cos^{-1}$ and domain of the function $\cos^{-1}$ is $\left [ -1, 1\right ]$.

Thus, $\cos ^{-1}:[-1,1] \rightarrow[0, \pi]$

Given :

$\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$

Explanation :

Let is first solve for  $\cos ^{-1} \frac{\sqrt{3}}{2}$.

Let,  $y=\cos ^{-1} \frac{\sqrt{3}}{2}$                                                         .....(i)

\begin{aligned} &\cos y=\frac{\sqrt{3}}{2} \\ &\cos y=\cos \frac{\pi}{6} \\ &y=\frac{\pi}{6} \\ &\cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6} \end{aligned}

$\because$ The range of principal value branch of $\cos^{-1}$ is $\left [ 0,\pi \right ]$.

$\therefore \; \; \; \; \; \; \; \; \quad \cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6} \in[0, \pi]$

The principal value of  $\cos ^{-1} \frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$                                                                         .....(ii)

Now,

\begin{aligned} &\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\} \\ &=\tan ^{-1}\left\{2 \sin \left(4 \times \frac{\pi}{6}\right)\right\} \end{aligned}                                                             [From equation (ii)]

$=\tan ^{-1}\left\{2 \sin \left(\frac{2 \pi}{3}\right)\right\}$

$=\tan ^{-1}\left(2 \times \frac{\sqrt{3}}{2}\right)$                                                                                  $\left[\because \sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}\right]$

$=\tan ^{-1}(\sqrt{3})$                                                                                                   .....(iii)

Now,

Let     $x=\tan ^{-1}(\sqrt{3})$                                                                                          ......(iv)

\begin{aligned} &\tan x=\sqrt{3} \\ &\tan x=\tan \frac{\pi}{3} \\ &x=\frac{\pi}{3} \end{aligned}

$\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$                                                                                      [From equation iv]

$\because$      The range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

$\because \quad \tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

$\therefore$    Principal value of ${\tan ^{-1}}(\sqrt{3})$ is $\frac{\pi}{3}$.

Hence principal vaue of  $\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$is $\frac{\pi}{3}$.