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provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise  3.7 question 4 sub question (viii)

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Answer: \frac{\pi }{6}

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{25 \pi}{6}\right)


First we solve \sec \frac{25 \pi}{6}

            \sec \left(\frac{25 \pi}{6}\right)=\sec \left(4 \pi+\frac{\pi}{6}\right)

As we know   \sec [2 \pi+\theta]=\sec \theta

            \sec \left(4 \pi+\frac{\pi}{6}\right)=\sec \left(\frac{\pi}{6}\right)

As we know  \sec \left(\frac{\pi}{6}\right)=\frac{2 \sqrt{3}}{3}

By substituting these value in \sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)  we get,

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

Now,     \text { let } y=\sec ^{-1}\left(\frac{2 \sqrt{3}}{3}\right)

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

            \begin{aligned} &\sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)=\frac{\pi}{6} \\ &\sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \end{aligned}


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