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3.Trigonometrical Ratios, Functions and Identities
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જો $A = 133^\circ ,$ તો $\;2\cos \frac{A}{2} = . . . .$
A
$ - \sqrt {1 + \sin A} - \sqrt {1 - \sin A} $
B
$ - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} $
C
$\sqrt {1 + \sin A} - \sqrt {1 - \sin A} $
D
$\sqrt {1 + \sin A} + \sqrt {1 - \sin A} $
Solution
(c) For $A = {133^o},\frac{A}{2} = {66.5^o}$
==> $\sin \frac{A}{2} > \cos \frac{A}{2} > 0$
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} $.
Standard 11
Mathematics
