JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Understanding +4 | -1

\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=2 \tan ^{-1} x

for

View answer and explanation

Correct option: C

|x| \leq 1

Option C (

|x| \leq 1

) is correct because the identity

\sin ^{-1}(\frac{2 x}{1+x^2})=2 \tan ^{-1} x

holds true only when

|x| \leq 1

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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},

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

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