Question

Let $△PQR$ be a right triangle, right angled at $R.$ If $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are the roots of the quadratic equation $a{x}^2+bx+c=0$, then which of the following is correct ?

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

Answer 1

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

We know that,
$P+Q+R=\pi \Rightarrow P+Q=\frac{\pi }{2}\Rightarrow \frac{P}{2}+\frac{Q}{2}=\frac{\pi }{4}\therefore \tan (\frac{P}{2}+\frac{Q}{2})=1\Rightarrow \frac{\tan \left(\frac{P}{2}\right)+\tan \left(\frac{Q}{2}\right)}{1-\tan \left(\frac{P}{2}\right)\tan \left(\frac{Q}{2}\right)}=1\cdots \left(1\right)$

As $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are the roots of the quadratic equation $a{x}^2+bx+c=0$

Now, sum of roots and product of roots is,
$\tan \left(\frac{P}{2}\right)+\tan \left(\frac{Q}{2}\right)=-\frac{b}{a}\tan \left(\frac{P}{2}\right)\tan \left(\frac{Q}{2}\right)=\frac{c}{a}$

Putting the values in the equation $\left(1\right)$,
$\frac{-\frac{b}{a}}{1-\frac{c}{a}}=1\Rightarrow \frac{-b}{a-c}=1\Rightarrow -b=a-c\therefore a+b=c$