Question

If $\sec \alpha$ and $cosec\alpha$ are the roots of the equation $x^2-px+q=0,$ then

• A. $p^2=p+2q$
• B. $q^2=p+2q$
• C. $p^2=q(q+2)$
• D. $q^2=p(p+2)$

Answer 1

Answer: (C)

$\sec \alpha +cosec\alpha =-\left(-p\right)=p$
$\text{Also}\sec \alpha cosec\alpha =q$
$\Rightarrow \sin \alpha \cos \alpha =\frac{1}{q}$
From (1) and (2)
$\sin \alpha +\cos \alpha =p\sin \alpha \cos \alpha =\frac{p}{q}$
squaring, we get
$1+2\sin \alpha \cos \alpha =\frac{p^2}{q^2}\Rightarrow 1+\frac{2}{q}=\frac{p^2}{q^2}$
$\Rightarrow p^2=q\left(q+2\right)$