JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

0 < x < \frac{\pi }{2},

then the minimum value of

\frac{{{{\cos }^3}x}}{{\sin x}} + \frac{{{{\sin }^3}x}}{{\cos x}}

View answer and explanation

Correct option: D

Option D (1) is correct because the minimum value is found using AM ≥ GM inequality, simplifying to

\sin 2x

, and then recognizing that the minimum value of

\sin 2x

in the given interval is 1 when

x = \pi/4

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)=