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) \[\sqrt{1+\sin A}+\sqrt{1-\sin A}\]
Correct Answer: C
Solution :
For \[A={{133}^{o}},\frac{A}{2}={{66.5}^{o}}\] Þ \[\sin \frac{A}{2}>\cos \frac{A}{2}>0\] 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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