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…

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
  1. $\frac{p^2}{p^2+q^2}$
  2. $\frac{p^2}{(p+q)^2}$
  3. $1-\frac{p}{(1-q)^2}$
  4. $\frac{p^2}{p^2+(1-q)^2}$

Solution

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 $\tan A \tan B=q$ $ \quad \begin{aligned} \therefore \quad \tan (A+B) & =\frac{\tan A+\tan B}{1-\tan A \tan B} \\ & =\frac{p}{1-q} \end{aligned}$ $\begin{aligned} & \Rightarrow \quad \sin (A+B)=\frac{p}{\sqrt{p^2+(1-q)^2}} \\ & \therefore \quad \sin ^2(A+B)=\frac{p^2}{p^2+(1-q)^2} \\ & \end{aligned}$

Asked in: AP EAMCET 2011

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