Question

In$∆PQR,\angle R=\frac{\mathrm{\pi }}{2}$, if $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are the roots of $a{x}^2+bx+c=0,a\ne 0$, then

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

Answer 1

The correct option is C

$c=a+b$

Explanation for the correct option:
Determining the resultant expression:

Given that $∆PQR,\angle R=\frac{\mathrm{\pi }}{2}$

$$\begin{gathered} \because \angle P+\angle Q+\angle R=\mathrm{\pi } \\ \Rightarrow \angle P+\angle Q+\frac{\mathrm{\pi }}{2}=\mathrm{\pi } \\ \Rightarrow \angle \mathrm{P}+\angle \mathrm{Q}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \frac{\angle \mathrm{P}}{2}+\frac{\angle \mathrm{Q}}{2}=\frac{\mathrm{\pi }}{4}.....\left(i\right)[multiplyingbothsidesby\frac{1}{2}] \end{gathered}$$

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

The Sum of roots of the quadratic equation is

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

Product of roots of the quadratic equation is

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

As we know

$\tan (A+B)=\frac{\tan \left(A\right)+\tan \left(A\right)}{1-\tan \left(A\right)\tan \left(B\right)}$

$$\begin{gathered} \therefore \tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\frac{\tan \left(\frac{P}{2}\right)+\tan \left(\frac{Q}{2}\right)}{1-\tan \left(\frac{P}{2}\right)\tan \left(\frac{Q}{2}\right)} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{4}\right)=\frac{\tan \left(\frac{P}{2}\right)+\tan \left(\frac{Q}{2}\right)}{1-\tan \left(\frac{P}{2}\right)\tan \left(\frac{Q}{2}\right)}\left[\text{from equation}\left(i\right)\&\tan \left(\frac{\mathrm{\pi }}{4}\right)=1\right] \\ \Rightarrow 1=\frac{{\displaystyle \frac{-b}{a}}}{1-{\displaystyle \frac{c}{a}}}[\text{from equations}(ii)\text{and}(iii\left)\right] \\ \Rightarrow 1=\frac{-b}{a-c} \\ \Rightarrow c=a+b \end{gathered}$$

Hence, the correct answer is option (C).