#### Need solution for RD Sharma maths Class 12 Chapter Inverse Trignometric Function Exercise 3.3 Question 3 Sub question (i) maths text book solution.

Answer : $\frac{3\pi}{4}$

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 $\cos^{-1}$.

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

The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch.

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

Domain of functions $\sin^{-1}$ and $\cos^{-1}$ are $\left [ -1,1 \right ]$ and $\left [ -1,1 \right ]$ respectively.

Given :    $\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$

Explanation :

Let us first solve for $\tan ^{-1}(1)$

Let

\begin{aligned} &y=\tan ^{-1}(1) \\ &\tan y=1 \end{aligned}                                                                              ...(i)

$\tan y=\tan \frac{\pi}{4} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \frac{\pi}{4}=1\right]$

\begin{aligned} &y=\frac{\pi}{4} \\ &\tan ^{-1}(1)=\frac{\pi}{4} \end{aligned}                                                                          [From equation (i)]

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

$\therefore \quad \tan ^{-1}(1)=\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Therefore principal of value of $\tan^{-1} (1)$ is  $\frac{\pi}{4}$                                                       ....(ii)

Now,

Let us solve for $\sin ^{-1}\left(\frac{-1}{2}\right)$

Let  $z=\sin ^{-1}\left(\frac{-1}{2}\right)$                                                                                      ....(iii)

$\begin{array}{ll} \sin z=\frac{-1}{2} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\because \sin \frac{\pi}{6}=\frac{1}{2}\right]} \\ \sin z=-\sin \frac{\pi}{6} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because \sin (-\theta)=-\sin \theta]} \end{array}$

\begin{aligned} &\sin z=\sin \left(\frac{-\pi}{6}\right) \\ &z=\frac{-\pi}{6} \\ &\sin ^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi}{6} \end{aligned}                                                                   [From equation (iii)]

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

$\because \sin ^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

$\therefore$ Principal value of $\sin ^{-1}\left(\frac{-1}{2}\right) \text { is } \frac{-\pi}{6}$                                                             ......(iv)

Now,

Let us solve for $\cos ^{-1}\left(\frac{-1}{2}\right)$

Let  $x=\cos ^{-1}\left(\frac{-1}{2}\right)$                                                                                            ....(v)

\begin{aligned} &\cos x=\frac{-1}{2} \\ &\cos x=-\cos \frac{\pi}{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned}

$\cos x=\cos \left(\pi-\frac{\pi}{3}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (\pi-\theta)=-\cos \theta]$

\begin{aligned} &\cos x=\cos \left(\frac{2 \pi}{3}\right) \\ &x=\frac{2 \pi}{3} \\ &\cos ^{-1}\left(\frac{-1}{2}\right)=\frac{2 \pi}{3} \end{aligned}                                                               [From equation (v)]

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

$\because \quad \cos ^{-1}\left(\frac{-1}{2}\right)=\frac{2 \pi}{3} \in[0, \pi]$

$\therefore$ Principal value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ is $\frac{2\pi}{3}$                                                                    ...(vi)

Now,

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

Putting the values from equations (ii), (iv) and (vi)

\begin{aligned} &=\frac{\pi}{4}+\frac{2 \pi}{3}+\left(\frac{-\pi}{6}\right) \\ &=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \\ &=\frac{3 \pi+8 \pi-2 \pi}{12} \\ &=\frac{9 \pi}{12} \\ &=\frac{3 \pi}{4} \end{aligned}