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

Answer : $\frac{\pi }{2}$

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$\cos ^{-1} \text { is }[-1,1]$.

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

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

Explanation :

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

Let  $y=\tan ^{-1}(-1)$                                                                              ....(i)

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

$\tan y=-\tan \frac{\pi}{4}$

$\tan y=\tan \left(\frac{-\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (-\theta)=-\tan \theta]$

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

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

\begin{aligned} &\tan ^{-1}(-1)=\frac{-\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\\ &\therefore \text { Principal value of }\\ &\tan ^{-1}(-1)=\frac{-\pi}{4} \end{aligned}                                                                            .....(ii)

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

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

$\cos x=\left(\frac{-1}{\sqrt{2}}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$

$\cos x=-\cos \frac{\pi}{4}$

$\cos x=\cos \left(\pi-\frac{\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (\pi-\theta)=-\cos \theta]$

\begin{aligned} &\cos x=\cos \left(\frac{3 \pi}{4}\right) \\ &x=\frac{3 \pi}{4} \\ &\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \end{aligned}                                                                            [From equation (iii)]

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

$\therefore \quad \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \in[0, \pi]$

$\therefore \text { Principal value of } \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right) \text { is } \frac{3 \pi}{4} \text { . }$                                                  .....(iv)

Now,

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

\begin{aligned} &=\frac{2 \pi}{4} \\ &=\frac{\pi}{2} \end{aligned}.