First slide

Theory of expressions

Question

Given that tan A and tan B are the roots of $x^{2} - p x + q = 0,$ the value of $\sin^{2} (A + B),$ is

Moderate

A

$\frac{p^{2}}{p^{2} + (1 - q)^{2}}$
B

$\frac{p^{2}}{p^{2} + q^{2}}$
C

$\frac{q^{2}}{p^{2} + (1 - q)^{2}}$
D

$\frac{p^{2}}{(p + q)^{2}}$

Solution

We have,
$\tan A + \tan B = p$ and $\tan A \tan B = q .$
$∴ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{p}{1 - q}$
Now,
$\sin^{2} (A + B) = \frac{1}{2} {1 - \cos 2 (A + B)}$
$\Rightarrow \sin^{2} (A + B) = \frac{1}{2} \{1 - \frac{1 - \tan^{2} (A + B)}{1 + \tan^{2} (A + B)}\} = \frac{\tan^{2} (A + B)}{1 + \tan^{2} (A + B)}$
$\Rightarrow \sin^{2} (A + B) = \frac{p^{2} / (1 - q)^{2}}{1 + \frac{p^{2}}{(1 - q)^{2}}} = \frac{p^{2}}{p^{2} + (1 - q)^{2}}$

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