JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

\tan ^{-1} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\alpha

, then

x^2

is equal to

View answer and explanation

Correct option: D

\sin 2 \alpha

Option D (

\sin 2 \alpha

) is correct because after applying componendo and dividendo and simplifying,

x^2

equals

\sin 2 \alpha

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x

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