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JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\cos ^2 48^{\circ}-\sin ^2 12^{\circ}=

View answer and explanation

Correct option: B

\frac{\sqrt{5}+1}{8}

Option B (

\frac{{\sqrt{5}+1}}{{8}}

) is correct because using the identity

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

, the expression simplifies to

\frac{{\sqrt{5}+1}}{{8}}

.

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

\alpha=6 \beta

\alpha

being an acute angle, then

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is equal to The value of

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

\triangle A B C

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C

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{\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)=