Question:
Write the value of $\tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)$.
Solution:
We know that $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$.
Now,
$\tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)=\tan ^{-1}\left(\frac{\frac{a}{b}-\frac{a-b}{a+b}}{1+\frac{a}{b} \frac{a-b}{a+b}}\right)$
$=\tan ^{-1}\left(\frac{\frac{a^{2}+a b-a b+b^{2}}{b(a+b)}}{\frac{a b+b^{2}-a b+a^{2}}{b(a+b)}}\right)$
$=\tan ^{-1}(1)$
$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$ $\left[\because \tan \frac{\pi}{4}=1\right]$
$=\frac{\pi}{4}$
$\therefore \tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)=\frac{\pi}{4}$