જો A = 133^circ , તો ;2cos frac{A}{2} = | vedclass

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Question

(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

Vedclass · Accepted Answer

$\sqrt {1 + \sin A} - \sqrt {1 - \sin A} $

Vedclass · Answer

(c) For $A = {133^o},\frac{A}{2} = {66.5^o}$

==> $\sin \frac{A}{2} > \cos \frac{A}{2} > 0$

Hence, $\sqrt {1 + \sin A} = \sin \frac{A}{2} + \cos \frac{A}{2}$&hellip;..$(i)$

and $\sqrt {1 - \sin A} = \sin \frac{A}{2} - \cos \frac{A}{2}$&hellip;..$(ii)$

Subtract $(ii)$ from $(i)$, $2\cos \frac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} $.

જો $A = 133^\circ ,$ તો $\;2\cos \frac{A}{2} = . . . .$

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