JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

The value of

\left\{\tan \left[\frac{\pi}{4}+\frac{1}{2} \sin ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \sin ^{-1}\left(\frac{a}{b}\right)\right]\right\}^{-1}

, where

0 < a < b

View answer and explanation

Correct option: C

\frac{\sqrt{b^2-a^2}}{2 b}

Option D (

\frac{\sqrt{b^2-a^2}}{2 b}

) is correct because simplifying the trigonometric expression and using the given condition leads to this result.

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

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{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)