Question

$\tan \alpha$ and $\tan \beta$ are the roots of the equation $x^2+ax+b=0$ then the value of ${\sin }^2(\alpha +\beta )+a\sin (\alpha +\beta )\cos (\alpha +\beta )+b{\cos }^2(\alpha +\beta )$ is equal to

• A. $ab$
• B. $b$
• C. $\frac{a}{b}$
• D. $a$

Answer 1

The correct option is B $b$

Given, $x^2+ax+b=0$

$\tan \alpha +\tan \beta =-a$

$\tan \alpha \tan \beta =b$

$\tan (\alpha +\beta )=\frac{-a}{1-b}=\frac{a}{b-1}$

${\sin }^2(\alpha +\beta )+a\sin (\alpha +beta)\cos (\alpha +\beta )+b{\cos }^2(\alpha +\beta )$

$=\frac{a^2}{a^2+(b-1{)}^2}+\frac{a\times (b-1)\times a}{a^2+(b-1{)}^2}+\frac{b\times (b-1{)}^2}{a^2+(b-1{)}^2}$

$=\frac{a^n+a^{2}b-a^2+b(b-1{)}^2}{a^2+(b-1{)}^2}$

$=\frac{b[a^2+(b-1{)}^2]}{a^2+(b-1{)}^2}$

$=b$.