Question

If in the triangle $ABC,C=\frac{\pi }{2}$ and $\tan \frac{A}{2},\tan \frac{B}{2}$ are two roots of the equation $p{x}^2+qx+r=0(p\ne 0)$ then show that $p+q=r$.

Answer 1

Sum of the roots$=\tan \frac{A}{2}+\tan \frac{B}{2}=\frac{-q}{p}$

Product of the roots$=\tan \frac{A}{2}\tan \frac{B}{2}=\frac{r}{p}$

In $△ABC,\angle A+\angle B+\angle C={180}^{\circ }$ by angle sum property

$\Rightarrow \angle A+\angle B={180}^{\circ }-\angle C$

$\Rightarrow \frac{\angle A+\angle B}{2}={90}^{\circ }-\frac{\angle C}{2}$

$\Rightarrow \tan \frac{\angle A}{2}+\tan \frac{+\angle B}{2}=\tan {90}^{\circ }-\frac{\pi }{4}$ where $\angle C=\frac{\pi }{2}$

$\Rightarrow \frac{\tan \frac{\angle A}{2}+\tan \frac{\angle B}{2}}{1-\tan \frac{\angle A}{2}\tan \frac{\angle B}{2}}=\cot \frac{\pi }{4}=1$

$\Rightarrow \tan \frac{\angle A}{2}+\tan \frac{\angle B}{2}=1-\tan \frac{\angle A}{2}\tan \frac{\angle B}{2}$

$\Rightarrow \tan \frac{\angle A}{2}+\tan \frac{\angle B}{2}=1-\tan \frac{\angle A}{2}\tan \frac{\angle B}{2}$

$\Rightarrow \frac{-q}{p}=1-\frac{r}{p}$

$\Rightarrow \frac{-q+r}{p}=1$

$\Rightarrow -q+r=p$

$\therefore p+q=r$

Hence proved.