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 the roots of the equation $a{x}^2+bx+c=0(a\ne 0)$ then

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

Answer 1

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

$a{x}^2+bc+c-0$, here $\tan \frac{P}{2}+\tan \frac{q}{2}-\frac{-b}{a}$ ...(1)
$\Rightarrow \tan \frac{p}{2}.\tan \frac{q}{2}-\frac{c}{a}$ ...(2)
Now $p+q-{90}^o\Rightarrow \frac{p}{2}+\frac{q}{2}-{45}^o$
$\Rightarrow \tan (\frac{p}{2}+\frac{q}{2})-\tan {45}^o$
$\Rightarrow \frac{\tan \frac{p}{2}+\tan \frac{q}{2}}{1-\tan \frac{p}{2}\tan \frac{q}{2}}-1$
$\Rightarrow \frac{-\frac{b}{a}}{1-\frac{c}{a}}-2\Rightarrow \frac{-b}{a}-1-\frac{c}{a}\Rightarrow a+b-c$