Question

If $\text{sec}\left(\text{α}\right)\text{and cosec}\left(\text{α}\right)$ are the roots of the equation $x^2+px+q=0$

• A. $p^2=p+2q$
• B. $q^2=p+2q$
• C. $p^2=q\left(q+2\right)$
• D. $q^2=p\left(p+2\right)$
• E. $p^2=q\left(q-2\right)$

Answer 1

The correct option is C

$p^2=q\left(q+2\right)$

Explanation for the correct option

Step 1: Information required for the solution

$\sec \left(\alpha \right)+\mathrm{cosec}\left(\alpha \right)=p\mathrm{and}\sec \left(\alpha \right)\times \mathrm{cosec}\left(\alpha \right)=q$

Step 2: Simplification and calculation for the given expression

We can write the expression as

Given equation is

$$\begin{gathered} x^2+px+q=0\left(\text{sec}\left(\text{α}\right)\text{and cosec}\left(\text{α}\right)\mathrm{are}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}\right) \\ \mathrm{Therefore},sec\alpha +\cos ec\alpha =p.....\left(1\right) \\ sec\alpha .\cos ec\alpha =q \\ \cos \alpha \times \sin \alpha =\frac{1}{q}\left(\sec \mathrm{\alpha }=\frac{1}{\cos \mathrm{\alpha }}\mathbf{and}\mathrm{cosec}\mathrm{\alpha }=\frac{1}{\sin \mathrm{\alpha }}\right) \\ \mathrm{From}\left(1\right)\mathrm{we}\mathrm{have} \\ \sin \alpha +\cos \alpha =p\cos \alpha \sin \alpha =\left(\frac{p}{q}\right).....\left(3\right) \\ \mathrm{Therefore},{\left(\sin \alpha +\cos \alpha \right)}^2=1+2\sin \alpha \cos \alpha \end{gathered}$$

Step 3: From equations (1), (2), and (3), we have

$$\begin{gathered} \therefore \frac{p^2}{q^2}=1+\frac{2}{q} \\ \Rightarrow p^2=q^2+2q \\ \Rightarrow p^2=q\left(q+2\right) \end{gathered}$$

Hence , the correct option is $\left(\mathrm{c}\right)$.