Question

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

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

Answer 1

Answer: (C)

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

Given $x^2+px+q=0(p\ne 0)$
having roots $\tan \alpha ,\tan \beta$
$\therefore \tan \alpha +\tan \beta =-p$
$tan\alpha \tan \beta =q$
$\therefore \tan (\alpha +\beta )=\frac{-p}{1-q}=\frac{p}{q-1}$