Question

If $\angle A=90^{°}$ in the $\Delta ABC$, then ${\tan }^{-1}\left(\frac{c}{a+b}\right)+{\tan }^{-1}\left(\frac{b}{a+c}\right)$ is equal to

• A. $0$
• B. $1$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (C)

We have, $ABC$ is a right angled $\Delta$ at $A$
$\Rightarrow a^2=b^2+c^2\dots \left(i\right)$
Now, ${\tan }^{-1}\left(\frac{c}{a+b}\right)+{\tan }^{-1}\left(\frac{b}{a+c}\right)$
$={\tan }^{-1}\left\{\frac{\left(\frac{c}{a+b}\right)+\left(\frac{b}{a+c}\right)}{1-\left(\frac{c}{a+b}\right)\left(\frac{b}{a+c}\right)}\right\}$

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