Question:
In a $\Delta A B C$, if $C$ is a right angle, then
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{4}$
(c) $\frac{5 \pi}{2}$
(d) $\frac{\pi}{6}$
Solution:
(b) $\frac{\pi}{4}$
We know
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\therefore \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=\tan ^{-1}\left(\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\frac{a}{b+c} \times \frac{b}{c+a}}\right)$
$=\tan ^{-1}\left(\frac{\frac{a c+a^{2}+b^{2}+b c}{(b+c)(c+a)}}{\frac{a c+c^{2}+b c}{(b+c)(c+a)}}\right)$
$=\tan ^{-1}\left(\frac{a c+c^{2}+b c}{a c+c^{2}+b c}\right) \quad\left[\because a^{2}+b^{2}=c^{2}\right]$
$=\tan ^{-1}(1)$
$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$
$=\frac{\pi}{4}$