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Q.

$\cos θ \cos (60^{\circ} + θ) \cdot \cos (60^{\circ} - θ) =$

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a

$\frac{1}{4} \tan 3 θ$

b

$\frac{1}{4} \cos 3 θ$

c

$\frac{1}{4} \sin 3 θ$

d

$\frac{1}{4} \cot 3 θ$

answer is B.

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Detailed Solution

$= \cos θ [\cos^{2} θ - \sin^{2} 60^{\circ}] = \cos θ [\cos^{2} θ - 3 / 4] = \frac{1}{4} \cos 3 θ$

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cos⁡θ cos⁡60∘+θ⋅cos⁡60∘−θ=