Answer
424.2k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)