If $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$, then $\sin^2(A+B)=$
- $\dfrac{p^2}{p^2+q^2}$
- $\dfrac{p^2}{(p+q)^2}$
- $1-\dfrac{p}{(1-q)^2}$
- $\dfrac{p^2}{p^2+(1-q)^2}$
Given tan A + tan B = p and tan A * tan B = q. Then tan(A+B) = (tan A + tan B) / (1 - tan A * tan B) = p / (1 - q). Using sin^2(theta) = tan^2(theta) / (1 + tan^2(theta)), we get sin^2(A+B) = (p/(1-q))^2 / (1 + (p/(1-q))^2) = p^2 / ((1-q)^2 + p^2).
Using the sum and product of roots formulas for the quadratic equation x^2 - px + q = 0, we have tan A + tan B = p and tan A tan B = q. The double angle formula for tangent gives tan(A+B) = (tan A + tan B) / (1 - tan A tan B) = p / (1 - q). Using the trigonometric identity, sin(A+B) equals tan(A+B) divided by the square root of (1 + tan^2(A+B)), so sin^2(A+B) = tan^2(A+B) / (1 + tan^2(A+B)). Substituting the value of tan(A+B) yields (p^2 / (1-q)^2) / (1 + p^2 / (1-q)^2), which simplifies to p^2 / ((1-q)^2 + p^2).