Question

Let in $\Delta ABC,\angle A=\frac{\pi }{2}$. Then value of ${\tan }^{-1}\frac{b}{a+c}+{\tan }^{-1}\frac{c}{a+b}$ equals

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{4}$

In $\Delta ABC$ ,

$\angle A={90}^{\circ }$

So, from Pythagorus theorem,

$a^2=b^2+c^2$ .............. (1)

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

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

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

$={\tan }^{-1}\frac{ab+ac+a^2}{a^2+ab+ac}$ from equation (1)

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

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