Question

The simplified form of ${\tan }^{-1}\frac{x}{y}-{\tan }^{-1}\frac{x-y}{x+y}$ is equal to

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{4}$

${\tan }^{-1}\frac{x}{y}-{\tan }^{-1}\frac{x-y}{x+y}$
$={\tan }^{-1}\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y}\times \frac{x-y}{x+y}}$
$={\tan }^{-1}\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}$
$={\tan }^{-1}\frac{x^2+xy-xy+y^2}{xy+y^2+x^2-xy}$
$={\tan }^{-1}\frac{x^2+y^2}{x^2+y^2}$
$={\tan }^{-1}1=\frac{\pi }{4}$