Question

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

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

Answer 1

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

Since ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$

Therefore,

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

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