Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

If $\sec α$ and $cosec α$ are the roots of $x^{2} - px + q = 0$ then,

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

p^{2} = q (q - 2)

p^{2} = q (q + 2)

p^{2} + q^{2} = 2 q

p^{2} + q^{2} = 1

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given equation is,

x^{2} - px + q

Given roots are

\sec α

and

cosec α

.
Here, product of roots is,

\sec α cosec α = \frac{Constant term}{coefficient of x^{2}}

\sec α cosec α = \frac{q}{1}

\Rightarrow \frac{1}{\sin α} \frac{1}{\cos α} = q

(Since,

\sec A = \frac{1}{\cos A}, cosec A = \frac{1}{\sin A}

)

\Rightarrow \frac{1}{\sin α \cos α} = q

\Rightarrow \sin α \cos α = \frac{1}{q}

So, the sum of roots is,

\Rightarrow \sec α + cosec α = - \frac{coefficient of x}{coefficient of x^{2}}

\Rightarrow \sec α + cosec α = - \frac{(- p)}{1}

\Rightarrow \frac{1}{\sin α} + \frac{1}{\cos α} = p

\Rightarrow \frac{\sin α + \cos α}{\sin α \cos α} = p

[

\sin α \cos α = \frac{1}{q}]

\Rightarrow \sin α + \cos α = \frac{p}{q}

Both sides square,

\Rightarrow {(\sin α + \cos α)}^{2} = {(\frac{p}{q})}^{2}

\Rightarrow \sin^{2} α + \cos^{2} α + 2 \sin α \cos α = \frac{p^{2}}{q^{2}}

\Rightarrow 1 + \frac{2}{q} = \frac{p^{2}}{q^{2}}

\Rightarrow p^{2} = q^{2} + 2 q

\Rightarrow p^{2} = q (q + 2)

This is the required relation.
Thus, the correct option is 2.

Watch 3-min video & get full concept clarity