The value of { ² 3 A}{ ² A}-{ ² 3 A}{ ² A} is

Correct Answer is : 8cos⁡2A8 2 A8cos2A

Correct Answer is : 8cos⁡2A8 2 A8cos2A

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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 104 of 150

Marks: +1, -0

The value of

\frac{\sin ^{2} 3 A}{\sin ^{2} A}-\frac{\cos ^{2} 3 A}{\cos ^{2} A}

Solution:

Concept:

Use triple-angle identities for

\sin 3A

and

\cos 3A

, then simplify using

\cos 2A

Explanation:

Write the triple-angle formulas:

\sin 3A = 3\sin A - 4\sin^3 A = \sin A(3 - 4\sin^2 A)

\cos 3A = 4\cos^3 A - 3\cos A = \cos A(4\cos^2 A - 3)

Substitute into the given expression:

\frac{\sin^2 3A}{\sin^2 A} - \frac{\cos^2 3A}{\cos^2 A} = (3 - 4\sin^2 A)^2 - (4\cos^2 A - 3)^2

Convert to

\cos 2A

using:

\sin^2 A = \frac{1 - \cos 2A}{2}, \quad \cos^2 A = \frac{1 + \cos 2A}{2}

Thus:

3 - 4\sin^2 A = 1 + 2\cos 2A

4\cos^2 A - 3 = 2\cos 2A - 1

So the expression becomes:

(1 + 2\cos 2A)^2 - (2\cos 2A - 1)^2

Using

a^2 - b^2 = (a-b)(a+b)

= [(1 + 2\cos 2A) - (2\cos 2A - 1)] \times [(1 + 2\cos 2A) + (2\cos 2A - 1)]

= 2 \times 4\cos 2A = 8\cos 2A

Answer:

8\cos 2A

, i.e., Option B.

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