The value of $\sin \,\,\left( {\pi + \theta } \right)\,\,\sin \,\,\left( {\pi - \theta } \right)\;{\rm{cose}}{{\rm{c}}^2}\,\theta $ is equal to

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

The value of

\sin \,\,\left( {\pi + \theta } \right)\,\,\sin \,\,\left( {\pi - \theta } \right)\;\mathrm{cose}{\mathrm{c}^2}\,\theta

is equal to

View answer and explanation

Correct option: A

 - 1

Option A (- 1) is correct because the expression simplifies to -1 using trigonometric identities.

Practice All Chapter MCQs →

\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)=