The value of ≤ft(60^{}-A) · A · ≤ft(60^{}+A) is

Correct Answer is : 14cos⁡3A1/4 3 A41cos3A

Correct Answer is : 14cos⁡3A1/4 3 A41cos3A

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MHTCET 16 Apr 2026 Shift 2 Paper

Question : 103 of 150

Marks: +1, -0

The value of

\cos \left(60^{\circ}-A\right) \cdot \cos A \cdot \cos \left(60^{\circ}+A\right)

Solution:

Concept:

Use the identity for the product of cosines and the triple-angle formula for cosine.

Explanation:

Let

E = \cos(60^\circ - A)\cos A \cos(60^\circ + A)

Use the identity

\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y

Here, take

x = 60^\circ

and

y = A

\cos(60^\circ - A)\cos(60^\circ + A) = \cos^2 60^\circ - \sin^2 A

Since

\cos 60^\circ = \frac{1}{2}

, we get

\cos^2 60^\circ = \frac{1}{4}

Thus,

E = \left(\frac{1}{4} - \sin^2 A\right)\cos A

Use

\sin^2 A = 1 - \cos^2 A

Then

E = \left(\frac{1}{4} - 1 + \cos^2 A\right)\cos A = \left(\cos^2 A - \frac{3}{4}\right)\cos A

E = \cos^3 A - \frac{3}{4}\cos A = \frac{1}{4}(4\cos^3 A - 3\cos A)

Using the identity

\cos 3A = 4\cos^3 A - 3\cos A

, we get

E = \frac{1}{4}\cos 3A

Answer:

Option A:

\frac{1}{4}\cos 3A

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