JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Medium

1 of 1

Application +4 | -1

{\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} - {\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}

, then the value of

x

View answer and explanation

Correct option: D

\frac{{a - b}}{{1 + ab}}

Option D (

\frac{{a - b}}{{1 + ab}}

) is correct because it is derived by using trigonometric identities to simplify the given equation and solve for

x

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →

\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)