Question

In a $\Delta$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. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{5\pi }{2}$

Answer 1

Answer: (B)

$\frac{\pi }{4}$

${\tan }^{-1}\left(\frac{a}{b+c}\right)+{\tan }^{-1}\left(\frac{b}{c+a}\right)$

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

Replace, $x$ by $\frac{a}{b+c}$ and $y$ by $\frac{b}{c+a}$

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

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

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

$={\tan }^{-1}\frac{a^2+b^2+bc+ac}{bc+c^2+ca}$

Given$:\angle C={90}^{\circ }\Rightarrow a^2+b^2=c^2$

$={\tan }^{-1}\frac{c^2+bc+ac}{bc+c^2+ca}$

$={\tan }^{-1}1=\frac{\pi }{4}$