Question

In a $\Delta ABC,A={90}^{\circ }$, then ${\tan }^{-1}\frac{b}{a+c}+{\tan }^{-1}\frac{c}{a+b}$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. ${\tan }^{-1}\frac{a}{b+c}$
• D. none of these

Answer 1

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

Given $\angle A={90}^0\therefore a^2=b^2+c^2$ (i)

$\therefore {\tan }^{-1}\frac{b}{a+c}+{\tan }^{-1}\frac{c}{a+b}={\tan }^{-1}\frac{\frac{b}{a+c}+\frac{c}{a+b}}{1-\frac{b}{a+c}.\frac{c}{a+b}}$

$={\tan }^{-1}\frac{ab+b^2+ac+c^2}{a^2+ac+ab+bc-bc}={\tan }^{-1}\frac{a^2+ab+ac}{a^2+ac+ab}={\tan }^{-1}1=\frac{\pi }{4}$