JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Analyzing +4 | -1

\sin \,\,x + {\sin ^2}x = 1

, then the value of

{\cos ^{12}}\,\,x + 3\,\,{\cos ^{10}}x + 3\,\,{\cos ^8}\,x + {\cos ^6}\,\,x + 2\,\,{\cos ^4}\,\,x + {\cos ^2}x - 2

, is equal to

View answer and explanation

Correct option: D

{\sin ^2}\,\,x

Option D (

{\sin ^2}\,\,x

) is correct because the given expression simplifies to

{\sin ^2}\,\,x

using the trigonometric identity and 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)=