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If trigonometric ratios $\sec \alpha $ and $\cos ec\alpha $ are the roots of the equation ${{x}^{2}}-px+q=0$ then
$\begin{align}
  & \text{A}\text{. }{{\text{p}}^{2}}+{{q}^{2}}=2q \\
 & \text{B}\text{. }{{\text{p}}^{2}}-{{q}^{2}}=2q \\
 & \text{C}\text{. }{{\text{p}}^{2}}+{{q}^{2}}=2p \\
 & \text{D}\text{. }{{\text{p}}^{2}}-{{q}^{2}}=2p \\
\end{align}$

Answer Verified Verified
Hint: We have given $\sec \alpha $ and $\cos ec\alpha $ are the roots of the equation ${{x}^{2}}-px+q=0$. We have to find the relation between the roots.
Now, we know that if $\alpha \text{ and }\beta $ are the roots of the equation $a{{x}^{2}}+bx+c=0$ then, the relation between the roots of the quadratic equation is given by
$\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$

Complete step-by-step solution:
We have given equation ${{x}^{2}}-px+q=0$ is a quadratic equation and $\sec \alpha $ and $\cos ec\alpha $ are roots of the equation.
So, the relation between $\sec \alpha $ and $\cos ec\alpha $will be
Sum of roots
$\begin{align}
  & \sec \alpha +\cos ec\alpha =\dfrac{-\left( -p \right)}{1} \\
 & \sec \alpha +\cos ec\alpha =p..............(i) \\
\end{align}$
Now, product of roots will be
\[\begin{align}
  & \sec \alpha .\cos ec\alpha =\dfrac{q}{1} \\
 & \sec \alpha .\cos ec\alpha =q \\
\end{align}\]
Now, we know that $\sec \alpha =\dfrac{1}{\cos \alpha }\text{ and cosec}\alpha \text{=}\dfrac{1}{\sin \alpha }\text{ }$
So, \[\begin{align}
  & \dfrac{1}{\cos \alpha }.\dfrac{1}{\sin \alpha }=q \\
 & \Rightarrow \cos \alpha .\sin \alpha =\dfrac{1}{q}................(ii) \\
\end{align}\]
Now, again consider equation (i)
$\sec \alpha +\cos ec\alpha =p$
Now, we know that $\sec \alpha =\dfrac{1}{\cos \alpha }\text{ and cosec}\alpha \text{=}\dfrac{1}{\sin \alpha }\text{ }$
Now, substitute the values in equation (i), we get
$\dfrac{1}{\cos \alpha }+\dfrac{1}{\sin \alpha }\text{= p }$
Now, solve further
$\begin{align}
  & \Rightarrow \dfrac{\sin \alpha +\cos \alpha }{\cos \alpha .\sin \alpha }\text{=p} \\
 & \Rightarrow \sin \alpha +\cos \alpha =p\cos \alpha .\sin \alpha \\
\end{align}$
Now, substitute the value from equation (ii), we get
$\Rightarrow \sin \alpha +\cos \alpha =\dfrac{p}{q}.............(iii)$
Now, we know that ${{\left( \sin \alpha +\cos \alpha \right)}^{2}}=1+2\sin \alpha .\cos \alpha $
${{\left( \sin \alpha +\cos \alpha \right)}^{2}}=1+2\sin \alpha .\cos \alpha $is derived from the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Here, $a=\sin \alpha $ and $b=\cos \alpha $ .
So, ${{\left( \sin \alpha +\cos \alpha \right)}^{2}}={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +2\sin \alpha .\cos \alpha $
We know that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$, so we get ${{\left( \sin \alpha +\cos \alpha \right)}^{2}}=1+2\sin \alpha .\cos \alpha $
Now, substituting the values from equation (i),(ii) and (iii), we get
$\begin{align}
  & \Rightarrow {{\left( \dfrac{p}{q} \right)}^{2}}=1+2\times \dfrac{1}{q} \\
 & \Rightarrow \dfrac{{{p}^{2}}}{{{q}^{^{2}}}}=1+\dfrac{2}{q} \\
 & \Rightarrow \dfrac{{{p}^{2}}}{{{q}^{^{2}}}}=\dfrac{q+2}{q} \\
 & \Rightarrow {{p}^{2}}=\dfrac{{{q}^{2}}\left( q+2 \right)}{q} \\
 & \Rightarrow {{p}^{2}}=q\left( q+2 \right) \\
 & \Rightarrow {{p}^{2}}={{q}^{2}}+2q \\
 & \Rightarrow {{p}^{2}}-{{q}^{2}}=2q \\
\end{align}$
Option B is the correct answer.

Note: In this question, we use the trigonometric identities. To solve this question we use the relation between the roots of the given quadratic equation because options are given like the relation between roots. Alternatively we can use the quadratic formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ but these will lead to lengthy solutions.