JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

Let

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

where

\left| x \right| < \frac{1}{{\sqrt 3 }}.

Then, a value of

y

View answer and explanation

Correct option: A

\frac{{3x - {x^3}}}{{1 - 3{x^2}}}

Option A (

\frac{{3x - {x^3}}}{{1 - 3{x^2}}}

) is correct because it is derived from the tangent addition formula and the double angle formula for tangent.

More Questions from INVERSE TRIGONOMETRY

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

\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)