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

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Answer:   \frac{\pi }{4}

Hint: The principal value branch of function  \cos ^{-1}  is  \left [ 0,\pi \right ]

Given:   \cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}


We know that \cos \left(-\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right) \quad \because[\cos (-\theta)=\cos \theta]                                                                           

Also know that, \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}

By substituting these values in \cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}

we get,

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

Let        y=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)

              \cos y=\frac{1}{\sqrt{2}}

Hence, range of principal value of \cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}

             \begin{aligned} &\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}=\frac{\pi}{4} \\ &\therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi] \end{aligned}

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