Solve
Question:

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

(A) $\frac{\pi}{2}$

(B). $\frac{\pi}{3}$

(C) $\frac{\pi}{4}$

(D) $\frac{-3 \pi}{4}$

Solution:

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

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

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

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

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

Hence, the correct answer is C

 

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