The value of ≤ft(18^{}-A) ≤ft(18^{}+A)- ≤ft(72^{}-A) ≤ft(72^{}+A) is equal to

Correct Answer is : sin⁡54∘ 54^{}sin54∘

Correct Answer is : sin⁡54∘ 54^{}sin54∘

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MHTCET 15 May 2024 Shift 1 Paper

Question : 104 of 150

Marks: +1, -0

The value of

\cos \left(18^{\circ}-A\right) \cos \left(18^{\circ}+A\right)-\cos \left(72^{\circ}-A\right) \cos \left(72^{\circ}+A\right)

is equal to

Solution:

Concept:

Use complementary angle identities:

\cos(90^\circ - \theta) = \sin \theta

and

\sin(X - Y) = \sin X \cos Y - \cos X \sin Y

Explanation:

Rewrite the angles

18^\circ + A

and

72^\circ + A

using complements:

18^\circ + A = 90^\circ - (72^\circ - A)

and

72^\circ + A = 90^\circ - (18^\circ - A)

Thus, the given expression becomes:

\cos(18^\circ - A)\sin(72^\circ - A) - \cos(72^\circ - A)\sin(18^\circ - A)

Rearranging this as

\sin(72^\circ - A)\cos(18^\circ - A) - \cos(72^\circ - A)\sin(18^\circ - A)

Using the identity

\sin(X - Y) = \sin X \cos Y - \cos X \sin Y

with

X = 72^\circ - A

and

Y = 18^\circ - A

, we get:

\sin[(72^\circ - A) - (18^\circ - A)] = \sin 54^\circ

Answer:

\sin 54^\circ

, which is option C.

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