JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 + x}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3},

then value of

x

View answer and explanation

Correct option: B

\frac{1}{{\sqrt 3 }}

Option B (

\frac{1}{{\sqrt 3 }}

) is correct because substituting

x = \tan \theta

and simplifying the equation leads to

x = \frac{1}{{\sqrt 3 }}

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{x^2} + {y^2} + {z^2} = {r^2},

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)