JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard

1 of 1

Application +4 | -1

ABCD

is a convex quadrilateral such that

4\sec A + 5 = 0

then the quadratic equation whose roots are

\tan A

and

\mathrm{cosec}\;A

View answer and explanation

Correct option: B

12{x^2} - 11x - 15 = 0

The correct answer is

12{x^2} - 11x - 15 = 0

because it is the quadratic equation formed with roots

\tan A

and

\csc A

derived from the given condition.

More Questions from TRIGONOMETRY RATIOS

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\tan \alpha=\frac{p}{q}

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=