Question

In a triangle $PQR,\angle R=\frac{\pi }{2}$. If $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are roots of $a{x}^2+bx+c=0$. where $a\ne 0$, then which one is true ?

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

Answer 1

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

We have,

$\tan \left(\frac{P}{2}\right).\tan \left(\frac{Q}{2}\right)aretherootsofa{x}^2+bx+c=0$

$\Rightarrow \tan \left(\frac{P}{2}\right)+\tan \left(\frac{Q}{2}\right)=-\frac{b}{a}$

$\Rightarrow \tan \left(\frac{P}{2}\right).\tan \left(\frac{Q}{2}\right)=\frac{c}{a}$

Now,

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

$\Rightarrow \frac{\frac{-b}{a}}{1-\frac{c}{a}}=1$

$\Rightarrow \frac{-b}{a}=\frac{a}{c}-\frac{c}{a}$

$\Rightarrow -b=a-c$

$\Rightarrow c=a+b$

Hence, this is the answer.