Question

In a ∆ ABC, if C is a right angle, then
${\tan }^{-1}\left(\frac{a}{b+c}\right)+{\tan }^{-1}\left(\frac{b}{c+a}\right)=$

(a) $\frac{\mathrm{\pi }}{3}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{5\mathrm{\pi }}{2}$

(d) $\frac{\mathrm{\pi }}{6}$

Answer 1

(b) $\frac{\mathrm{\pi }}{4}$

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

$$\begin{gathered} \therefore {\tan }^{-1}\left(\frac{a}{b+c}\right)+{\tan }^{-1}\left(\frac{b}{c+a}\right)={\tan }^{-1}\left(\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\frac{a}{b+c}\times \frac{b}{c+a}}\right) \\ ={\tan }^{-1}\left(\frac{\frac{ac+a^2+b^2+bc}{\left(b+c\right)\left(c+a\right)}}{\frac{ac+c^2+bc}{\left(b+c\right)\left(c+a\right)}}\right) \end{gathered}$$

$$\begin{gathered} ={\tan }^{-1}\left(\frac{ac+c^2+bc}{ac+c^2+bc}\right)\left[\because a^2+b^2=c^2\right] \\ ={\tan }^{-1}\left(1\right) \\ ={\tan }^{-1}\left(\tan \frac{\mathrm{\pi }}{4}\right) \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$