^ - 1 ( x y ) - ^ - 1 , ( x - y x + y ) is

MCQ

${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\,\left( {\frac{{x - y}}{{x + y}}} \right)$ is

A
$\frac{\pi }{2}$
B
$\frac{\pi }{3}$
✓
$\frac{\pi }{4}$
D
$\frac{\pi }{4} $ or $ - \frac{{3\pi }}{4}$

✓

Answer

Correct option: C.

$\frac{\pi }{4}$

c
(c) ${\tan ^{ - 1}}\frac{x}{y} - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right) = {\tan ^{ - 1}}\frac{x}{y} - {\tan ^{ - 1}}\left( {\frac{{1 - y/x}}{{1 + y/x}}} \right)$

$ = {\tan ^{ - 1}}\frac{x}{y} - \left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\frac{y}{x}} \right)$

$ = {\tan ^{ - 1}}\frac{x}{y} + {\tan ^{ - 1}}\frac{y}{x} - \frac{\pi }{4}$

$ = {\tan ^{ - 1}}\frac{x}{y} + {\cot ^{ - 1}}\frac{x}{y} - \frac{\pi }{4} = \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}$.

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