Question

In a triangle $PQR,\angle R=\frac{\pi }{2}$. If $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of the equation $a{x}^2+bx+c=0(a\ne 0)$ , then

• A. $a=b+c$
• B. $c=a+b$
• C. $b=c$
• D. $b=a+c$

Answer 1

The correct option is A $a=b+c$

We know that,

$\tan \left(\frac{P+Q}{2}\right)=\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2}\tan \frac{Q}{2}}$ ..... $\left(i\right)$

Since, $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of the equation $a{x}^2+bx+c=0(a\ne 0)$

$\Rightarrow \tan \frac{P}{2}+\tan \frac{Q}{2}=\frac{-b}{a}$ and

$\tan \frac{P}{2}\tan \frac{Q}{2}=\frac{c}{a}$

In $△PQR,\angle P+\angle Q+\angle R=\pi$

$\Rightarrow \angle P+\angle Q=\pi -\angle R$
$\tan \left(\frac{\pi -R}{2}\right)=\frac{\frac{-b}{a}}{1-\frac{c}{a}}$ ...... From $\left(i\right)$
$\Rightarrow \cot \frac{R}{2}=\frac{b}{c-a}$

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