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MHTCET 13 Apr 2026 Shift 1 Paper

Question : 115 of 150

Marks: +1, -0

\triangle A B C,(b-c)^{2} \cos ^{2} \frac{A}{2}+(b+c)^{2} \sin ^{2} \frac{A}{2}=

Solution:

Concept:

Use half-angle identities and the cosine rule to simplify the expression.

Explanation:

\triangle ABC

, the half-angle formulas are:

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

Substitute these into the given expression:

(b-c)^2 \cos^2 \frac{A}{2} + (b+c)^2 \sin^2 \frac{A}{2} = \frac{1}{2}\left[(b-c)^2(1+\cos A) + (b+c)^2(1-\cos A)\right].

Group the terms containing

\cos A

= \frac{1}{2}\left[(b-c)^2+(b+c)^2 + \cos A\left((b-c)^2-(b+c)^2\right)\right].

Now,

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

and

(b-c)^2-(b+c)^2 = -4bc

So the expression becomes:

= \frac{1}{2}\left[2b^2+2c^2 - 4bc\cos A\right] = b^2 + c^2 - 2bc\cos A.

By the cosine rule in

\triangle ABC

a^2 = b^2 + c^2 - 2bc\cos A.

Therefore:

(b-c)^2 \cos^2 \frac{A}{2} + (b+c)^2 \sin^2 \frac{A}{2} = a^2.

Answer:

Option A:

a^2

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