MCQ
$\frac{1}{4}\left(\sqrt{3} \cos 23^{\circ}-\sin 23^{\circ}\right)=$
- A$\cos 43^{\circ}$
- B$\cos 7^{\circ}$
- C$\frac{1}{4} \cos 53^{\circ}$
- ✓$\frac{1}{2} \cos 53^{\circ}$
✓
Answer
Correct option: D.
$\frac{1}{2} \cos 53^{\circ}$
(D)
$\frac{1}{4}\left(\sqrt{3} \cos 23^{\circ}-\sin 23^{\circ}\right)$
$-\frac{1}{2}\left(\cos 30^{\circ} \cos 23^{\circ}-\sin 30^{\circ} \sin 23^{\circ}\right)$$\ldots\left[\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\frac{1}{2}\right]$
$=\frac{1}{2} \cos \left(30^{\circ}+23^{\circ}\right)=\frac{1}{2} \cos 53^{\circ}$
$\frac{1}{4}\left(\sqrt{3} \cos 23^{\circ}-\sin 23^{\circ}\right)$
$-\frac{1}{2}\left(\cos 30^{\circ} \cos 23^{\circ}-\sin 30^{\circ} \sin 23^{\circ}\right)$$\ldots\left[\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\frac{1}{2}\right]$
$=\frac{1}{2} \cos \left(30^{\circ}+23^{\circ}\right)=\frac{1}{2} \cos 53^{\circ}$
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