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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
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Verified Answer
The correct answer is:
$\frac{p^2}{p^2+(1-q)^2}$
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}$
$\begin{aligned} & \Rightarrow \quad \sin (A+B)=\frac{p}{\sqrt{p^2+(1-q)^2}} \\ & \therefore \quad \sin ^2(A+B)=\frac{p^2}{p^2+(1-q)^2} \\ & \end{aligned}$
$\therefore \tan A+\tan B=p$ 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}$
$\begin{aligned} & \Rightarrow \quad \sin (A+B)=\frac{p}{\sqrt{p^2+(1-q)^2}} \\ & \therefore \quad \sin ^2(A+B)=\frac{p^2}{p^2+(1-q)^2} \\ & \end{aligned}$
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