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Question: For A = \(133^{o},2\cos\frac{A}{2}\) is equal to...

For A = 133o,2cos⁡A2133^{o},2\cos\frac{A}{2} is equal to

A

−1+sin⁡A−1−sin⁡A- \sqrt{1 + \sin A} - \sqrt{1 - \sin A}

B

−1+sin⁡A+1−sin⁡A- \sqrt{1 + \sin A} + \sqrt{1 - \sin A}

C

1+sin⁡A−1−sin⁡A\sqrt{1 + \sin A} - \sqrt{1 - \sin A}

D

1+sin⁡A+1−sin⁡A\sqrt{1 + \sin A} + \sqrt{1 - \sin A}

Answer

1+sin⁡A−1−sin⁡A\sqrt{1 + \sin A} - \sqrt{1 - \sin A}

Explanation

Solution

For A=133o,A2=66.5oA = 133^{o},\frac{A}{2} = 66.5^{o} ⇒ sin⁡A2cos⁡A2\sin\frac{A}{2\cos\frac{A}{2}}

Hence, 1+sin⁡A=sin⁡A2+cos⁡A2\sqrt{1 + \sin A} = \sin\frac{A}{2} + \cos\frac{A}{2} ......(i) and

1−sin⁡A=sin⁡A2−cos⁡A2\sqrt{1 - \sin A} = \sin\frac{A}{2} - \cos\frac{A}{2} ......(ii)

Subtract (ii) from (i) we get, 2cos⁡A2=1+sin⁡A−1−sin⁡A2\cos\frac{A}{2} = \sqrt{1 + \sin A} - \sqrt{1 - \sin A}.