If $x \geq 1$, then $2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^2}$ is equal to

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

x \geq 1

, then

2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^2}

is equal to

View answer and explanation

Correct option: B

\pi

Option B (

\pi

) is correct because when x ≥ 1, the expression 2tan⁻¹x + sin⁻¹(2x/(1+x²)) simplifies to π.

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x

{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)