Question

If $\Delta ABC$ is right angled at $C$, then ${\tan }^{-1}\left(\frac{a}{b+c}\right)+{\tan }^{-1}\left(\frac{b}{c+a}\right)=$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{4}$

Here, $k={\tan }^{-1}\left(\frac{a}{b+c}\right)+ta{n}^{-1}\left(\frac{b}{a+c}\right)$

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

$k={\tan }^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{a+c}}{1-\frac{ab}{(a+c)(b+c)}}\right]$

$={\tan }^{-1}\left(\frac{a^2+ac+b^2+bc}{ab+ac+bc+c^2-ab}\right)$

By cosine rule for triangle
$\cos C=\frac{a^2+b^2-c^2}{2bc}$

$={\tan }^{-1}\left(\frac{2bc\cos c+c^2+ac+bc}{ac+bc+c^2}\right)$

Given $C={90}^o,\cos C=0$

$={\tan }^{-1}\left(\frac{c^2+ac+bc}{ac+bc+c^2}\right)$

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

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