Q. 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) $q^{2} = p(p + 2)$
Explanation
(c)
Here, $\sec{\alpha + cosec\ \alpha = \ p}\$and$\sec{\alpha.cosec\ \alpha = q}$
$\Rightarrow \ \frac{\sin{\alpha + \cos\alpha}}{\sin{\alpha\cos\alpha}} = p$ and $\sin{\alpha\cos{\alpha = \frac{1}{q}}}$
$\Rightarrow \left( \sin{\alpha + \cos\alpha} \right)^{2} = \left( \frac{p}{q} \right)^{2}$
$\Rightarrow \sin^{2}{\alpha + \cos^{2}{\alpha + 2\sin{\alpha\cos{\alpha = \frac{p^{2}}{q^{2}}}}}}$
$\Rightarrow q^{2}\left( 1 + \frac{2}{q} \right) = p^{2}$
$\Rightarrow q(q + 2) = p^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.