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  • explain solution rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 5 sub question (iv) maths

explain solution rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 5 sub question (iv) maths

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Answer:{\frac{-\pi }{6}}

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

Given:    \sin^{-1}\left ( \frac{-\sqrt{3}}{2} \right )+\cos^{-1}\left ( \frac{\sqrt{3}}{2} \right )

Solution:

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

Since \sin ^{-1}x is an angle in  \left ( \frac{-\pi }{2},\frac{\pi }{2} \right ); \; \cos ^{-1}xis an angle in \left ( 0,\pi \right )

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

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