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