Q. Given that $\tan A$ and $\tan B$ are the roots of $x^{2} - px + q = 0$, then the value of $\sin^{2}{(A + B)}$ is
A) $\frac{p^{2}}{p^{2}(1 - q)^{2}}$
B) $\frac{q^{2}}{p^{2} + q^{2}}$
C) $\frac{q^{2}}{p^{2} - (1 - q^{2})}$
D) $\frac{p^{2}}{p^{2} + q^{2}}$
Explanation
(a)
Here, $\tan{A + \tan{B = p}}$ and $\tan{A\tan{B = q}}$
Now, $\tan{(A + B) = \frac{\tan{A + \tan B}}{1 - \tan{A\tan B}} = \frac{p}{1 - q}}$
$\therefore\ \sin^{2}{(A + B) = \frac{1 - \cos{\lbrack 2(A + B)\rbrack}}{2}}$
$= \frac{1}{2}\left\lbrack 1 - \frac{1 - \tan^{2}{(A + B)}}{1 + \tan^{2}{(A + B)}} \right\rbrack$
$= \frac{1}{2}\left\lbrack 1 - \frac{1 - \left( \frac{p}{1 - q} \right)^{2}}{1 + \left( \frac{p}{1 - q} \right)^{2}} \right\rbrack$
$= \frac{1}{2}\left\lbrack \frac{(1 - q)^{2} + p^{2} - (1 - q)^{2} + p^{2}}{(1 - q)^{2} + p^{2}} \right\rbrack$
$= \frac{p^{2}}{p^{2} + (1 - q)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.