Find the value

Question:

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

Solution:

We know 

$\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}, x y>-1$
Now,

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

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

$=\tan ^{-1}\left\{\frac{\frac{x^{2}+x y-x y+y^{2}}{y(x+y)}}{\frac{x^{2}+y^{2}+x y-x y}{3(x+y)}}\right\}$

$=\tan ^{-1} 1$

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

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

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

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