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

Given :  $\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$

Explanation :

\begin{aligned} &\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \\ &=\tan ^{-1}\left\{\tan \left(\pi-\frac{\pi}{6}\right)\right\}+\cos ^{-1}\left\{\cos \left(2 \pi+\frac{\pi}{6}\right)\right\}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (2 \pi+\theta)=\cos \theta] \end{aligned}

\begin{aligned} &=\tan ^{-1}\left\{-\left(\tan \frac{\pi}{6}\right)\right\}+\cos ^{-1}\left\{\cos \left(\frac{\pi}{6}\right)\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (\pi-\theta)=-\tan \theta] \\ &=\tan ^{-1}\left\{\tan \left(\frac{-\pi}{6}\right)\right\}+\frac{\pi}{6} \\ &=\frac{-\pi}{6}+\frac{\pi}{6} \\ &=0 \end{aligned}

$\therefore \quad \tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=0$