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) $