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explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (vi) maths

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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 \left(-\frac{7 \pi}{3}\right)\right\}


First we solve   \sec \left(-\frac{7 \pi}{3}\right)

As we know that   \sec (-\theta)=\sec (\theta)

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

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

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

            \sec ^{-1}(2)

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

            \begin{aligned} &\text { 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}

Hence,  \sec ^{-1}\left(\sec -\frac{7 \pi}{3}\right)=\frac{\pi}{3}


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