JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

x \ge 1

, then

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

is equal to

View answer and explanation

Correct option: D

\pi

Option D is correct because for

x \ge 1

{\sin ^{ - 1}}(\frac{{2x}}{{1 + {x^2}}}) = \pi - 2{\tan ^{ - 1}}x

, and thus the expression simplifies to

\pi

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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)