Question

In $\Delta ABC$ with the length of sides as $a,b,c,$ if ${\tan }^{-1}\left(\frac{b}{c+a}\right)+{\tan }^{-1}\left(\frac{c}{a+b}\right)=\frac{\pi }{4}$ and $a+b+c=2s$, then its area is

• A. $s(s-a)$
• B. $s(s-b)$
• C. $s(s-c)$
• D. $s^2$

Answer 1

The correct option is A $s(s-a)$

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

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{b}{c+a}+\frac{c}{a+b}}{1-\left(\frac{b}{c+a}\right)\left(\frac{c}{a+b}\right)}\right]=\frac{\pi }{4}$

$\Rightarrow \frac{b(a+b)+c(c+a)}{(c+a)(a+b)-bc}=\tan \frac{\pi }{4}=1$

$\Rightarrow ab+b^2+c^2+ca=ca+bc+a^2+ab-bc$
$\Rightarrow b^2+c^2=a^2$
$\Rightarrow \angle A={90}^{\circ }$

We know that $\cot \frac{A}{2}=\frac{s(s-a)}{\Delta }$
$\Rightarrow \cot {45}^{\circ }=\frac{s(s-a)}{\Delta }$
$\Rightarrow \Delta =s(s-a)$