MCQ
For $A = 133^\circ ,\;2\cos \frac{A}{2}$ is equal to
- A$ - \sqrt {1 + \sin A} - \sqrt {1 - \sin A} $
- B$ - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} $
- ✓$\sqrt {1 + \sin A} - \sqrt {1 - \sin A} $
- D$\sqrt {1 + \sin A} + \sqrt {1 - \sin A} $
==> $\sin \frac{A}{2} > \cos \frac{A}{2} > 0$
Hence, $\sqrt {1 + \sin A} = \sin \frac{A}{2} + \cos \frac{A}{2}$…..$(i)$
and $\sqrt {1 - \sin A} = \sin \frac{A}{2} - \cos \frac{A}{2}$…..$(ii)$
Subtract $(ii)$ from $(i)$, $2\cos \frac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} $.
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