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Question: If \(\tan A = \frac{1 - \cos B}{\sin B},\) find\(\tan 2A\)in terms of \(\tan B\)and show that...

If tan⁡A=1−cos⁡Bsin⁡B,\tan A = \frac{1 - \cos B}{\sin B}, findtan⁡2A\tan 2Ain terms of tan⁡B\tan Band show that

A

tan⁡2A=tan⁡B\tan 2A = \tan B

B

tan⁡2A=tan⁡2B\tan 2A = \tan^{2}B

C

tan⁡2A=tan⁡2B+2tan⁡B\tan 2A = \tan^{2}B + 2\tan B

D

None of the above

Answer

tan⁡2A=tan⁡B\tan 2A = \tan B

Explanation

Solution

tan⁡A=1−cos⁡Bsin⁡B=2sin⁡2(B/2)2sin⁡(B/2)cos⁡(B/2)=tan⁡B2\tan A = \frac{1 - \cos B}{\sin B} = \frac{2\sin^{2}(B/2)}{2\sin(B/2)\cos(B/2)} = \tan\frac{B}{2}

⇒ tan⁡2A=tan⁡B\tan 2A = \tan B.