Question

7 In $△ABC$ if $\angle A=\frac{\pi }{2}$, then ${\tan }^{-1}\frac{b}{c+a}+{\tan }^{-1}\frac{c}{a+b}$ equals

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{7\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

${\tan }^{-1}\frac{b}{c+a}+{\tan }^{-1}\frac{c}{a+b}={\tan }^{-1}\left(\frac{\frac{b}{c+a}+\frac{c}{a+b}}{1-\frac{bc}{(c+a)(a+b)}}\right)$
$={\tan }^{-1}\left(\frac{ab+ca+b^2+c^2}{ca+ab+a^2}\right)={\tan }^{-1}1=\frac{\pi }{4}$