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Provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise  3.7 question 2 sub question (iv)

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Answer:   \frac{\pi }{6}

Hint: The principal value branch of function \cos ^{-1} is \left [ 0,\pi \right ]
Given:  \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)


First we solve \cos \frac{13 \pi}{6}

            \begin{aligned} &\cos \frac{13 \pi}{6}=\cos \left(2 \pi+\frac{\pi}{6}\right) \\ &\therefore[\cos (2 \pi+\theta)]=\cos \theta \\ &\cos \left(2 \pi+\frac{\pi}{6}\right)=\cos \left(\frac{\pi}{6}\right) \end{aligned}

            \begin{gathered} =\frac{\sqrt{3}}{2} \\ \cos \left(\frac{13 \pi}{6}\right)=\frac{\sqrt{3}}{2} \end{gathered}

By substituting these values in \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)

we get,

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

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

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

Hence, range of principal value of \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}

            \begin{aligned} &\cos ^{-1}\left(\cos \frac{\pi}{6}\right)=\frac{\pi}{6} \\ &\therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi] \\ &\cos ^{-1}\left\{\cos \left(\frac{13 \pi}{6}\right)\right\}=\frac{\pi}{6} \end{aligned}


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