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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

The value of

\sin ^{-1}\left[\cos \left\{\cos ^{-1}(\cos x)+\sin ^{-1}(\sin x)\right\}\right]

; where

x \in\left(\frac{\pi}{2}, \pi\right)

is

View answer and explanation

Correct option: D

\frac{-\pi}{2}

Option D (

\frac{-\pi}{2}

) is correct because, for

x \in (\frac{\pi}{2}, \pi)

,

\sin^{-1}(\sin x) = \pi - x

and

\cos^{-1}(\cos x) = x

, leading to

\sin^{-1}(\cos(\pi)) = \frac{-\pi}{2}

.

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\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

{x^2} + {y^2} + {z^2} = {r^2},

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)