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#### Provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise  3.7 question 4 sub question (iv)

Answer: $\frac{\pi }{3}$

Hint: The range of principal value of  $\sec ^{-1}$ is $\left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]$
Given:  $\sec ^{-1}\left(\sec \frac{7 \pi}{3}\right)$

Explanation:

First we solve $\sec \frac{7 \pi}{3}$

\begin{aligned} &\sec \left(\frac{7 \pi}{3}\right)=\sec \left(2 \pi+\frac{\pi}{3}\right) \\ &\therefore[\sec (2 \pi+\theta)=\sec \theta] \end{aligned}

\begin{aligned} &\sec \left(2 \pi+\frac{\pi}{3}\right)=\sec \left(\frac{\pi}{3}\right) \\ &\therefore \sec \left(\frac{\pi}{3}\right)=2 \end{aligned}

By substituting these value in $\sec ^{-1}\left(\sec \frac{7 \pi}{3}\right)$ we get,

$\sec ^{-1}(2)$

Now,     $\text { let } y=\sec ^{-1}(2)$

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

The range of principal value of   $\sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{\pi}{3}\right)=2$

\begin{aligned} &\sec ^{-1}\left(\sec \frac{\pi}{3}\right)=\frac{\pi}{3} \\ &\sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \end{aligned}