Question

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

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

Answer 1

The correct option is D $\frac{p^2}{p^2+(1-q{)}^2}$

Since $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$, we get

$\tan A+\tan B=p$

$\tan A\times \tan B=q$

$\Rightarrow \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\times \tan B}$

$\Rightarrow \tan (A+B)=\frac{p}{1-q}$

$\Rightarrow \sin (A+B)=\frac{p}{\sqrt{p^2+(1-q{)}^2}}$

$\Rightarrow {\sin }^2(A+B)=\frac{p^2}{p^2+(1-q{)}^2}$