JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

Let

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

such that

\tan ^{-1}\left(\frac{\tan \alpha}{3+2 \tan ^2 \alpha}\right)+\tan ^{-1}\left(\frac{2 \tan \alpha}{3}\right)=\frac{\pi}{12}

, then

\alpha

equals:

View answer and explanation

Correct option: D

\frac{\pi}{12}

Option D (

\frac{\pi}{12}

) is correct because by simplifying the given equation using trigonometric identities, we find that

\alpha

equals

\frac{\pi}{12}

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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},

{\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)