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Question: \[\frac{\sin 2A}{1 + \cos 2A}.\frac{\cos A}{1 + \cos A} =\]...

sin⁡2A1+cos⁡2A.cos⁡A1+cos⁡A=\frac{\sin 2A}{1 + \cos 2A}.\frac{\cos A}{1 + \cos A} =

A

tan⁡A2\tan\frac{A}{2}

B

cot⁡A2\cot\frac{A}{2}

C

sec⁡A2\sec\frac{A}{2}

D

cosecA2\text{cosec}\frac{A}{2}

Answer

tan⁡A2\tan\frac{A}{2}

Explanation

Solution

(sin⁡2A1+cos⁡2A)(cos⁡A1+cos⁡A)\left( \frac{\sin 2A}{1 + \cos 2A} \right)\left( \frac{\cos A}{1 + \cos A} \right)

=2sin⁡Acos⁡A2cos⁡2Acos⁡A1+cos⁡A=sin⁡A1+cos⁡A=tan⁡A2= \frac{2\sin A\cos A}{2\cos^{2}A}\frac{\cos A}{1 + \cos A} = \frac{\sin A}{1 + \cos A} = \tan\frac{A}{2}.