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JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)=

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Correct option: A

\frac{1}{2} \cos 2 \theta

Option A (

\frac{{1}}{{2}} \cos 2 \theta

) is correct because using the identity

\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)

, the expression simplifies to

\frac{{1}}{{2}} \cos 2 \theta

.

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

\alpha=6 \beta

\alpha

being an acute angle, then

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

is equal to The value of

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

\triangle A B C

is acute angled at

C

, then The value of the expression

{\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)=