Question

If tan A, tan B are the roots of $x^2-Px+Q=0$ the value of $si{n}^2$ (A+B)=(where P, Q $ϵ$ R)

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

Answer 1

The correct option is B

$\frac{P^2}{P^2+(1-Q{)}^2}$

$tan(A+B)=\frac{P}{1-Q}thensi{n}^2(A+B)=\frac{ta{n}^2(A+B)}{1+ta{n}^2(A+B)}$