Question

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

• A. ${\sin }^2(\alpha +\beta )+p\sin (\alpha +\beta )\cos (\alpha +\beta )+q{\cos }^2(\alpha +\beta )=q$
• B. $\tan (\alpha +\beta )=\frac{p}{q-1}$
• C. $\cos (\alpha +\beta )=1-q$
• D. $\sin (\alpha +\beta )=-p$

Answer 1

Answer: (A), (B)

The correct options are
A ${\sin }^2(\alpha +\beta )+p\sin (\alpha +\beta )\cos (\alpha +\beta )+q{\cos }^2(\alpha +\beta )=q$
B $\tan (\alpha +\beta )=\frac{p}{q-1}$

Hence
$\tan \alpha +\tan \beta =-p$
And
$\tan \alpha .\tan \beta =q$
Now
$\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }$
$=\tan (\alpha +\beta )$
$=\frac{-p}{1-q}$

$=\frac{p}{q-1}$.
Hence
$\sin (\alpha +\beta )=p$.
$\cos (\alpha +\beta )=q-1$.
Hence
$p^2+p\left(p\right)(q-1)+q(q-1{)}^2$

$=p^2+p^2(q-1)+q(q-1{)}^2$
$=p^2.q+q(q-1{)}^2$
$=q(p^2+(q-1{)}^2)$
$=q\left(\sin \right(\alpha +\beta {)}^2+\cos (\alpha +\beta {)}^2)$
$=q$.