If $\sin ^{-1} x=\tan ^{-1} y$, what is the value of $\frac{1}{x^2}-\frac{1}{y^2}$ ?

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

\sin ^{-1} x=\tan ^{-1} y

, what is the value of

\frac{1}{x^2}-\frac{1}{y^2}

View answer and explanation

Correct option: A

Option A (1) is correct because by substituting

\sin^{-1}x = \tan^{-1}y = \theta

and using trigonometric identities, the expression simplifies to 1.

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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)