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. $b+c=a$
• C. $a+c=b$
• D. $b=c$

Answer 1

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

Given In $△PQR,\angle R=\frac{\pi }{2}$

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

$\because \angle R=\frac{\pi }{2},\angle P+\angle Q=\pi -\angle R$

$\angle P+\angle Q=\frac{\pi }{2}$

$\frac{\angle P}{2}+\frac{\angle Q}{2}=\frac{\pi }{4}$

Apply tangent on both sides

$\tan [\frac{\angle P}{2}+\frac{\angle Q}{2}]=\tan \left(\frac{\pi }{4}\right)$

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

From the equation $a{x}^2+bx+c$ where $\tan \frac{P}{2}$ and

$\tan \frac{Q}{2}$ are roots

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

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

Substitute above equation in $\left(I\right)$

$|tan(\frac{\angle P}{2}+\frac{\angle Q}{2})=1$

$\Rightarrow \frac{\tan \frac{\angle P}{2}+\tan \frac{\angle Q}{2}}{1-\tan \frac{\angle P}{2}.\tan \frac{\angle Q}{2}}=1$ $[\because \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }]$

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

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

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

$\Rightarrow c-b=a$

$\Rightarrow c=a+b$

$a+b=c$