JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

The value of

\cos \left(270^{\circ}+\theta\right) \cdot \cos \left(90^{\circ}-\theta\right)-

\sin \left(270^{\circ}-\theta\right) \cos \theta

View answer and explanation

Correct option: D

Option D (1) is correct because after applying trigonometric identities and simplifying, the expression reduces to

\sin^2 \theta + \cos^2 \theta

, which equals 1.

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

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\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)=