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Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (vii) maths textbook solution

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Answer: \frac{3\pi }{4}

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{13 \pi}{4}\right)


First we solve  \sec \left(\frac{13 \pi}{4}\right)

            \sec \left(\frac{13 \pi}{4}\right)=\sec \left(4 \pi-\frac{3 \pi}{4}\right)

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

But here after subtract \frac{3\pi }{4}   it comes in III quadrant where sec is negative

            \begin{aligned} &\sec \left(4 \pi-\frac{3 \pi}{4}\right)=-\sec \left(\frac{3 \pi}{4}\right) \\ &\therefore \sec \frac{3 \pi}{4}=\sqrt{2} \\ &-\sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}

By substituting these values in \sec ^{-1}\left(\sec \frac{13 \pi}{4}\right) we get,

            \sec ^{-1}(-\sqrt{2})

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

            \begin{aligned} &\sec y=-\sqrt{2} \\ &-\sec y=\sqrt{2} \\ &-\sec \left(\frac{\pi}{4}\right)=\sqrt{2} \\ &\sec \left(\pi-\frac{\pi}{4}\right)=\sqrt{2} \\ &\sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}

The range of the principle value of   \sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\} \text { and } \sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2}

            \sec ^{-1}\left(\sec \frac{3 \pi}{4}\right)=\frac{3 \pi}{4}

            \therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}


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