Mark the tick against the correct answer in the following: The value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ is A. $\frac{13 \pi}{6}$ B. $\frac{5 \pi}{6}$ C. $\frac{\pi}{6}$ D. $\frac{7 \pi}{6}$

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Mark the tick against the correct answer in the following: The value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ is A. $\frac{13 \pi}{6}$ B. $\frac{5 \pi}{6}$ C. $\frac{\pi}{6}$ D. $\frac{7 \pi}{6}$

CBSE | Class 12 | Inverse Trigonometric Functions | Maths | RS Aggarwal

Mark the tick against the correct answer in the following: The value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ is A. $\frac{13 \pi}{6}$ B. $\frac{5 \pi}{6}$ C. $\frac{\pi}{6}$ D. $\frac{7 \pi}{6}$

Solution:

Option(C) ids correct.
To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$
Now, let $x=\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$
$\Rightarrow \cos x=\cos \left(\frac{13 \pi}{6}\right)$
Here, range of principle value of $\cos$ is $[0, \pi]$
$\Rightarrow x=\frac{13 \pi}{6} \notin[0, \pi]$
Hence for all values of $x$ in range $[0, \pi]$ , the value of $\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$ is
$\begin{array}{l} \Rightarrow \cos x=\cos \left(2 \pi-\frac{\pi}{6}\right)\left(\because \cos \left(\frac{13 \pi}{6}\right)=\cos \left(2 \pi-\frac{\pi}{6}\right)\right) \\ \Rightarrow \cos x=\cos \left(\frac{\pi}{6}\right)(\because \cos (2 \pi-\theta)=\cos \theta) \\ \Rightarrow x=\frac{\pi}{6} \end{array}$