Question

The value of $ta{n}^{-1}\left(\frac{x}{y}\right)-ta{n}^{-1}\left(\frac{x-y}{x+y}\right),x,y>0$ is

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

Answer 1

Answer: (A)

$\frac{\pi }{4}$

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

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

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

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

$={\tan }^{-1}\left(1\right)$
$={45}^0$

$=\frac{\pi }{4}$.