Q51 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
If $\cos 43^\circ + \sin 43^\circ = k^3$, then $\cos 2^\circ = \ldots$
- A$-\dfrac{k^3}{\sqrt{2}}$
- Correct answerB$\dfrac{k^3}{\sqrt{2}}$
- C$-\dfrac{k^3}{\sqrt{3}}$
- D$\dfrac{k^3}{\sqrt{3}}$
Solution
Given $\cos 43^\circ + \sin 43^\circ = k^3$
Multiplying and dividing the LHS by $\sqrt{2}$:
$\sqrt{2} \left( \dfrac{1}{\sqrt{2}} \cos 43^\circ + \dfrac{1}{\sqrt{2}} \sin 43^\circ \right) = k^3$
$\sqrt{2} (\sin 45^\circ \cos 43^\circ + \cos 45^\circ \sin 43^\circ) = k^3$
Using the identity $\sin A \cos B + \cos A \sin B = \sin(A + B)$:
$\sqrt{2} \sin(45^\circ + 43^\circ) = k^3$
$\sqrt{2} \sin 88^\circ = k^3$
Since $\sin \theta = \cos(90^\circ - \theta)$:
$\sqrt{2} \cos(90^\circ - 88^\circ) = k^3$
$\sqrt{2} \cos 2^\circ = k^3$
$\cos 2^\circ = \dfrac{k^3}{\sqrt{2}}$
Answer: $\dfrac{k^3}{\sqrt{2}}$
Multiplying and dividing the LHS by $\sqrt{2}$:
$\sqrt{2} \left( \dfrac{1}{\sqrt{2}} \cos 43^\circ + \dfrac{1}{\sqrt{2}} \sin 43^\circ \right) = k^3$
$\sqrt{2} (\sin 45^\circ \cos 43^\circ + \cos 45^\circ \sin 43^\circ) = k^3$
Using the identity $\sin A \cos B + \cos A \sin B = \sin(A + B)$:
$\sqrt{2} \sin(45^\circ + 43^\circ) = k^3$
$\sqrt{2} \sin 88^\circ = k^3$
Since $\sin \theta = \cos(90^\circ - \theta)$:
$\sqrt{2} \cos(90^\circ - 88^\circ) = k^3$
$\sqrt{2} \cos 2^\circ = k^3$
$\cos 2^\circ = \dfrac{k^3}{\sqrt{2}}$
Answer: $\dfrac{k^3}{\sqrt{2}}$
Hence, the correct answer is option: 'B'.