Question

If $\tan \alpha ,\tan \beta$ are the roots of the equation $x^2+px+q=0(p\ne 0)$ then
${\sin }^2(\alpha +\beta )+p\sin (\alpha +\beta )\cos (\alpha +\beta )+q{\cos }^2(\alpha +\beta )=$

• A. $0$
• B. $1$
• C. $p$
• D. $q$

Answer 1

The correct option is D $q$

$x^2+px+q=0$
$\tan \alpha +\tan \beta =-p$
$\tan \alpha \tan \beta =q$
$\tan (\alpha +\beta )=\frac{-p}{1-q}=\frac{p}{q-1}$
$\therefore {\sin }^2(\alpha +\beta )+p\sin (\alpha +\beta )\cos (\alpha +\beta )+q{\cos }^2(\alpha +\beta )$
${\cos }^2(\alpha +\beta )\left[{\tan }^2\right(\alpha +\beta )+p\tan (\alpha +\beta )+q]$
$=[\frac{p^2}{(q-1{)}^2}+\frac{p^2}{q-1}+1]{\cos }^2(\alpha +\beta )$
$=\left[\frac{p^2+p^{2}q-p^2+q(q^2-2p+1)}{(q-1{)}^2}\right]{\cos }^2(\alpha +\beta )$
$=\left[\frac{p^{2}q+q^3-2pq+q}{(q-1{)}^2}\right]{\cos }^2(\alpha +\beta )$
$=q\frac{(p^2+q^2-2p+1)}{(q-1{)}^2}\frac{(q-1{)}^2}{(p^2+q^2-2q+1)}$
$=q$