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Please solve rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 5 sub question (i) maths textbook solution

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Answer:{\frac{2\pi }{3}}

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

Given:    \cos ^{-1}\left ( \frac{1}{2} \right )+2\sin ^{-1}\left ( \frac{1}{2} \right )

Solution:

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

        \cos x=\frac{1}{2}=\cos \left ( {\frac{\pi }{3}}\right )

      \cos ^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{3}              ....(1)

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

       \sin y=\frac{1}{2}=\sin \left ( {\frac{\pi }{6}}\right )

     \sin ^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{6}              .....(2)

From (1) and (2),

     \cos ^{-1}\left ( \frac{1}{2} \right )+2\sin ^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{3}+2\left ( \frac{\pi }{6} \right )

                                                        =\frac{\pi }{3}+\frac{\pi }{3}

                                                        =\frac{2\pi }{3}

\thereforePrincipal value of \cos ^{-1}\left ( \frac{1}{2} \right )+2\sin ^{-1}\left ( \frac{1}{2} \right )=\frac{2\pi }{3}

 

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