Question

In a triangle ABC,$\angle C=\frac{\pi }{2}.$ If $\tan \left(\frac{A}{2}\right)$and $\tan \left(\frac{B}{2}\right)$are the roots of the equation $a{x}^2+bx+c=0(a\ne 0)$ then the value of$\frac{a+b}{c}$ (where a,b,c,are sides of opposite to angles A,B,C,respectively )is

Answer 1

Given in $\Delta ABC,\angle C=\frac{\pi }{2}$
$\Rightarrow A+B=\frac{\pi }{2}$
$\Rightarrow \frac{A+B}{2}=\frac{\pi }{4}$
Also given $\tan \left(\frac{A}{2}\right)$and $\tan \left(\frac{B}{2}\right)$are the roots of the equation $a{x}^2+bx+c=0(a\ne 0)$
$\Rightarrow \tan \frac{A}{2}\tan \frac{B}{2}=\frac{c}{a}$
Also, $\tan \frac{A}{2}+\tan \frac{B}{2}=-\frac{b}{a}$
$\Rightarrow \tan \frac{(A+B)}{2}[1-\tan \frac{A}{2}\tan \frac{B}{2}]=-\frac{b}{a}$

$\Rightarrow [1-\tan \frac{A}{2}\tan \frac{B}{2}]=-\frac{b}{a}$
$\Rightarrow 1-\frac{c}{a}=-\frac{b}{a}$
$\Rightarrow a+b=c$
$\Rightarrow \frac{a+b}{c}=1$