First slide

Theory of expressions

Question

In a triangle $A B C, C = \frac{π}{2}$ and $\tan \frac{A}{2}, \tan \frac{B}{2}$ are the roots of the equation $p x^{2} + q x + r = 0,$ $p \neq 0$ then

Moderate

A

$p + q = r$
B

$q + r = 0$
C

$p + r = q$
D

$q = r .$

Solution

The given quadratic equation is $p x^{2} + q x + r = 0$
$\tan \frac{A}{2}, \tan \frac{B}{2}$ are the given roots equation
Sum of the roots $\tan \frac{A}{2} + \tan \frac{B}{2} = \frac{- q}{p}$
Product of the roots $\tan \frac{A}{2} . \tan \frac{B}{2} = \frac{r}{p}$
Given $C = \frac{π}{2}$
$\Rightarrow A + B = π - C = π - \frac{π}{2} = \frac{π}{2}$
$\Rightarrow \frac{A}{2} + \frac{B}{2} = \frac{π}{2}$
Applying $\tan$ on both sides
$\Rightarrow \tan (\frac{A}{2} + \frac{B}{2}) = 1$ $(∵ \tan \frac{π}{2} = 1)$
$\Rightarrow \frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = 1$ $(∵ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B})$
$\Rightarrow \frac{- \frac{q}{p}}{1 - \frac{r}{p}} = 1$
$\Rightarrow \frac{- q}{p - r} = 1 \Rightarrow p + q = r .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App