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MHTCET 11 Apr 2026 Shift 2 Paper

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Question : 104 of 150

Marks: +1, -0

The value of

\cos ^{2} 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^{2} 50^{\circ}

is equal to....

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$
1

Solution:

Concept:

Use sum-to-product identities for

\cos^2 A + \cos^2 B

and

\cos A \cos B

to simplify the expression.

Explanation:

Let

A = 10^\circ

and

B = 50^\circ

.

Using

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

:

\cos^2 10^\circ + \cos^2 50^\circ = 1 + \cos 60^\circ \cos 40^\circ = 1 + \frac{1}{2}\cos 40^\circ

.

Using

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

:

\cos 10^\circ \cos 50^\circ = \frac{1}{2}[\cos 60^\circ + \cos 40^\circ] = \frac{1}{4} + \frac{1}{2}\cos 40^\circ

.

Now substitute into the given expression

E

:

E = \left(1 + \frac{1}{2}\cos 40^\circ\right) - \left(\frac{1}{4} + \frac{1}{2}\cos 40^\circ\right)

.

The

\frac{1}{2}\cos 40^\circ

terms cancel, leaving

E = 1 - \frac{1}{4} = \frac{3}{4}

.

Answer:

\frac{3}{4}

, i.e. Option C.

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