${\tan ^{ - 1}}\frac{x}{y} - {\tan ^{ - 1}}\frac{{x - y}}{{x + y}}\;$ is equal to (where $x 0)$

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

{\tan ^{ - 1}}\frac{x}{y} - {\tan ^{ - 1}}\frac{{x - y}}{{x + y}}\;

is equal to

(where

x < y > 0)

View answer and explanation

Correct option: B

\frac{\pi }{4}

Option B (

\frac{\pi }{4}

) is correct because by simplifying the expression using the tangent subtraction formula and the identity

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

, it simplifies to

\frac{\pi}{4}

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