Please solve RD Sharma Class 12 Chapter Inverse Trigonometric Functions Exercise 3.3 Question 1 Subquestion (iii) maths textbook solution.

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

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

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

Explanation :

Let  $y=\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$                                                                                                  ....(i)

$y=\tan ^{-1}(0) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{2}=0\right]$

$\tan \; y=0$

$\tan y=\tan 0^{\circ} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan 0^{\circ}=0\right]$

$y=0$

$\tan ^{-1}\left(\cos \frac{\pi}{2}\right)=0$

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

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

Hence, the principal value of $\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$ is 0.