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