Question

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

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\mathrm{\pi }$

Answer 1

The correct option is B

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

Find the value of :

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

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

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

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

Hence, Option ‘B’ is Correct.