JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

The value of

\left(1+\cos \frac{\pi}{6}\right)\left(1+\cos \frac{\pi}{3}\right)\left(1+\cos \frac{2 \pi}{3}\right)\left(1+\cos \frac{7 \pi}{6}\right)

\frac{m}{16}

. Value of

m

View answer and explanation

Correct option: A

Option A (6) is correct because given \cos θ =

\frac{{-3}}{{5}}

and θ in the third quadrant, calculating the expression results in

\frac{{1}}{{6}}

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)=