#### Provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise  3.7 question 2 sub question (iv)

Answer:   $\frac{\pi }{6}$

Hint: The principal value branch of function $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given:  $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$

Explanation:

First we solve $\cos \frac{13 \pi}{6}$

\begin{aligned} &\cos \frac{13 \pi}{6}=\cos \left(2 \pi+\frac{\pi}{6}\right) \\ &\therefore[\cos (2 \pi+\theta)]=\cos \theta \\ &\cos \left(2 \pi+\frac{\pi}{6}\right)=\cos \left(\frac{\pi}{6}\right) \end{aligned}

$\begin{gathered} =\frac{\sqrt{3}}{2} \\ \cos \left(\frac{13 \pi}{6}\right)=\frac{\sqrt{3}}{2} \end{gathered}$

By substituting these values in $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$

we get,

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

Now,     $\operatorname{let} y=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

\begin{aligned} &\cos y=\frac{\sqrt{3}}{2} \\ &\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2} \end{aligned}

Hence, range of principal value of $\cos ^{-1} \text { is }[0, \pi] \text { and } \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$

\begin{aligned} &\cos ^{-1}\left(\cos \frac{\pi}{6}\right)=\frac{\pi}{6} \\ &\therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi] \\ &\cos ^{-1}\left\{\cos \left(\frac{13 \pi}{6}\right)\right\}=\frac{\pi}{6} \end{aligned}