If tan A and tan B are the roots of the quadr | vedclass

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(D) Given that $	an A$ and $	an B$ are the roots of the quadratic equation $x^2-px+q=0$. By the relation between roots and coefficients,we have: $

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

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(D) Given that $	an A$ and $	an B$ are the roots of the quadratic equation $x^2-px+q=0$. By the relation between roots and coefficients,we have: $	an A + 	an B = p$ $	an A 	an B = q$ Using the trigonometric identity for $	an(A+B)$: $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{p}{1-q}$ Now,we know that $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$. Substituting $	an(A+B) = \frac{p}{1-q}$: $\sin^2(A+B) = \frac{(\frac{p}{1-q})^2}{1 + (\frac{p}{1-q})^2} = \frac{\frac{p^2}{(1-q)^2}}{\frac{(1-q)^2 + p^2}{(1-q)^2}} = \frac{p^2}{p^2+(1-q)^2}$

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

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