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

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

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

\sec \alpha

and

\mathrm{cosec}\;\alpha

are the roots of the equation

{x^2} - px + q = 0

, then

View answer and explanation

Correct option: C

{p^2} = q\left( {q + 2} \right)

Option C (

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

) is correct because by using the sum and product of roots, and the trigonometric identity

\sin^2 \alpha + \cos^2 \alpha = 1

, we can derive the given relation.

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