Question

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

 

 

Options

• ${\displaystyle \frac{\pi }{3}}$

• ${\displaystyle \frac{\pi }{4}}$

• ${\displaystyle \frac{5x}{2}}$

• ${\displaystyle \frac{\pi }{6}}$

Answer 1

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

We know
$${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$$
$$\therefore {\tan }^{-1}\left(\frac{a}{b+c}\right)+{\tan }^{-1}\left(\frac{b}{c+a}\right)={\tan }^{-1}\left(\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\frac{a}{b+c}\times \frac{b}{c+a}}\right)$$
$$={\tan }^{-1}\left(\frac{\frac{ac+a^2+b^2+bc}{(b+c)(c+a)}}{\frac{ac+c^2+bc}{(b+c)(c+a)}}\right)$$
$$={\tan }^{-1}\left(\frac{ac+c^2+bc}{ac+c^2+bc}\right)[\because a^2+b^2=c^2]$$
$$={\tan }^{-1}\left(1\right)$$
$$={\tan }^{-1}(\tan \frac{\pi }{4})$$
$$=\frac{\pi }{4}$$