JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

\sin ^{-1}\left(\frac{1}{5}\right)+\sec ^{-1}(2)+2 \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sec ^{-1}(5)

+\sin ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(\sqrt{3})={k} \pi

, then

{k}=

View answer and explanation

Correct option: A

Option A (2) is correct because by using trigonometric identities and simplifying, the value of k is found to be 2.

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\sin \left(2 \tan ^{-1}(1 / 3)\right)+

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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)