Question

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

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

Answer 1

Answer: (C)

$c=a+b$

if $\angle R=\frac{\pi }{2};\angle Q=\frac{\pi }{2}-\angle P$

Sum of the roots $=\frac{-b}{a}=\tan \left(\frac{P}{2}\right)+\tan (\frac{\pi }{4}-\frac{P}{2})=\frac{1-\tan \frac{P}{2}}{1+\tan \left(\frac{P}{2}\right)}+\tan \left(\frac{P}{2}\right)=\frac{1+{\tan }^2\left(\frac{P}{2}\right)}{1+\tan \left(\frac{P}{2}\right)}$

Product of the roots $\frac{c}{a}=\tan \left(\frac{P}{2}\right)\times \tan (\frac{\pi }{4}-\frac{P}{2})$

$=\frac{1-\tan \frac{P}{2}}{1+\tan \frac{P}{2}}\times \tan \frac{P}{2}$

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

$a=1+\tan \left(\frac{P}{2}\right)$

$b=-1-{\tan }^2\left(\frac{P}{2}\right)$

$c=\tan \left(\frac{P}{2}\right)-{\tan }^2\left(\frac{P}{2}\right)$

From the given equations, we can see $c=a+b$