If 43^{}+ 43^{}=k³, then 2^{}=.

Correct Answer is : k32{k³}{{2}}2k3

Correct Answer is : k32{k³}{{2}}2k3

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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 103 of 150

Marks: +1, -0

\cos 43^{\circ}+\sin 43^{\circ}=k^{3}

\cos 2^{\circ}=\ldots

Solution:

Concept:

Use the identity

\cos A + \sin A = \sqrt{2}\cos(45^{\circ}-A)

to express the given sum in terms of

\cos 2^{\circ}

Explanation:

We are given

\cos 43^{\circ} + \sin 43^{\circ} = k^{3}

Apply the standard identity with

A = 43^{\circ}

So,

\cos 43^{\circ} + \sin 43^{\circ} = \sqrt{2}\cos(45^{\circ} - 43^{\circ})

Simplify the angle inside the cosine:

45^{\circ} - 43^{\circ} = 2^{\circ}

Thus,

\cos 43^{\circ} + \sin 43^{\circ} = \sqrt{2}\cos 2^{\circ}

From the given condition, this equals

k^{3}

Therefore,

k^{3} = \sqrt{2}\cos 2^{\circ}

Divide both sides by

\sqrt{2}

to isolate

\cos 2^{\circ}

This gives

\cos 2^{\circ} = \frac{k^{3}}{\sqrt{2}}

Answer:

\cos 2^{\circ} = \frac{k^{3}}{\sqrt{2}}

, so the correct option is B.

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