JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

x_1=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right), x_2=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)

; where

x \in

(0,1)

then

x_1+x_2

is equal to

View answer and explanation

Correct option: C

\pi

Option C (

\pi

) is correct because simplifying the expression using trigonometric identities and the given condition leads to

\pi

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

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

\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)