Question

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

• A. $0$
• B. $1$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{6}$
• E. $\frac{\mathrm{\pi }}{8}$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{4}$

Find the value of :

${\tan }^{-1}\left(\frac{c}{a+b}\right)+{\tan }^{-1}\left(\frac{b}{a+c}\right)$:

Given, $\angle A=90°$ in the triangle ABC

$\Rightarrow$$a^2=b^2+c^2$

$\begin{array}{rcl}\therefore {\tan }^{-1}\left(\frac{c}{a+b}\right)+{\tan }^{-1}\left(\frac{b}{a+c}\right)& =& {\tan }^{-1}\left[\frac{\left(\frac{c}{a+b}\right)+\left(\frac{b}{a+c}\right)}{1-\left(\frac{c}{a+b}\right)\left(\frac{b}{a+c}\right)}\right]\\ & =& ta{n}^{-1}\left(\frac{ac+c^2+b^2+ab}{a^2+ab+ac}\right)\\ & =& {\tan }^{-1}\left(\frac{ac+a^2+ab}{a^2+ab+ac}\right)\\ & =& {\tan }^{-1}\left(1\right)\\ & =& \frac{\mathbf{\pi }}{\mathbf{4}}\end{array}$

Hence, Option ‘C’ is Correct.