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

Answer: (B)

We have, ${\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\left(\frac{x-y}{x+y}\right)$
$={\tan }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}\left(\frac{y-x}{y+x}\right)$
$={\tan }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}\left(\frac{1-x/y}{1+x/y}\right)$
$={\tan }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}(1)-{\tan }^{-1}\left(\frac{x}{y}\right)={\tan }^{-1}(1)$
$\left[\because {\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)\right]$
$=\frac{\pi }{4}$