Question

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^{2}-p x+q=0$, then $\sin ^{2}(A+B)$ is equal to

• A. $\frac{p^{2}}{p^{2}+q^{2}}$
• B. $\frac{p^{2}}{(p+q)^{2}}$
• C. $1-\frac{p}{(1-q)^{2}}$
• D. $\frac{p^{2}}{p^{2}+(1-q)^{2}}$

Answer 1

Answer: (D)

Since, $\tan A$ and $\tan B$ are the roots of the equation $x^{2}-p x+q=0$
$\therefore \tan A+\tan B=p$ and $\tan A \tan B=q$
$\therefore \tan (A+B) =\frac{\tan A+\tan B}{1-\tan A \tan B} $
$=\frac{p}{1-q} $
$ \Rightarrow \sin (A+B) =\frac{p}{\sqrt{p^{2}+(1-q)^{2}}} $
$\therefore \sin ^{2}(A+B)=\frac{p^{2}}{p^{2}+(1-q)^{2}}$