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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

\tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)} \right]\;

is equal to

View answer and explanation

Correct option: C

\frac{{2a}}{{1 - {a^2}}}

Option C (

\frac{{2a}}{{1 - {a^2}}}

) is correct because the expression simplifies to

\tan(2{\tan ^{ - 1}}a)

, which equals

\frac{{2a}}{{1 - {a^2}}}

.

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\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

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\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\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)