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MHTCET 18 Apr 2026 Shift 1 Paper

Question : 103 of 150

Marks: +1, -0

\triangle A B C

\angle C=\frac{\pi}{3}

\cos ^{2} A+\cos ^{2} B+\cos A \cos B

is...

Solution:

Concept:

Use the triangle angle sum and standard trigonometric identities to simplify the given expression.

Explanation:

\triangle ABC

, we have

A + B + C = \pi

Given

C = \frac{\pi}{3}

, we get:

A + B = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.

Use the identities:

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

and

\cos A \cos B = \frac{\cos(A+B) + \cos(A-B)}{2}.

So the expression becomes:

\cos^2 A + \cos^2 B + \cos A \cos B = 1 + \cos(A+B)\cos(A-B) + \frac{\cos(A+B) + \cos(A-B)}{2}.

Now,

\cos(A+B) = \cos\frac{2\pi}{3} = -\frac{1}{2}

Substitute this value:

= 1 - \frac{1}{2}\cos(A-B) + \frac{-\frac{1}{2} + \cos(A-B)}{2}.

Simplifying:

= 1 - \frac{1}{2}\cos(A-B) - \frac{1}{4} + \frac{1}{2}\cos(A-B) = 1 - \frac{1}{4} = \frac{3}{4}.

Thus, the required value is

\frac{3}{4}

Answer:

\frac{3}{4}

Option A.

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