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

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

\mathrm{\;}\sec \,\,\mathrm{\alpha \;}\,\,\mathrm{and\;}\,\,\mathrm{cosec}\,\,\mathrm{\alpha \;}

are the root of the equation

{x^2} - px + q = 0

,then

View answer and explanation

Correct option: B

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

Option A (

\mathrm{1}

) is correct because by substituting

\sin x = {\cos ^2}x

into the expression and simplifying, we obtain 1.

Practice All Chapter MCQs →

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{x^2} - px + q = 0

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x

a

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