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Question: If \(\frac{\cos A}{\cos B} = \frac{\sin(C - \theta)}{\sin(C + \theta)}\), then \(\tan\theta\)is equa...

If cos⁡Acos⁡B=sin⁡(C−θ)sin⁡(C+θ)\frac{\cos A}{\cos B} = \frac{\sin(C - \theta)}{\sin(C + \theta)}, then tan⁡θ\tan\thetais equals

A

tan⁡(A+B)2tan⁡(A−B)2tan⁡C2\tan\frac{(A + B)}{2}\tan\frac{(A - B)}{2}\tan\frac{C}{2}

B

tan⁡(A+B)2tan⁡(A−B)2tan⁡C\tan\frac{(A + B)}{2}\tan\frac{(A - B)}{2}\tan C

C

tan⁡(A+B)2tan⁡(A−B)2sin⁡C2\tan\frac{(A + B)}{2}\tan\frac{(A - B)}{2}\sin\frac{C}{2}

D

tan⁡(A+B)2tan⁡(A−B)2cos⁡C2\tan\frac{(A + B)}{2}\tan\frac{(A - B)}{2}\cos\frac{C}{2}

Answer

tan⁡(A+B)2tan⁡(A−B)2tan⁡C\tan\frac{(A + B)}{2}\tan\frac{(A - B)}{2}\tan C

Explanation

Solution

cos⁡Acos⁡B=sin⁡(C−θ)sin⁡(C+θ)\frac{\cos A}{\cos B} = \frac{\sin(C - \theta)}{\sin(C + \theta)}

⇒ cos⁡A−cos⁡Bcos⁡A+cos⁡B=sin⁡(C−θ)−sin⁡(C+θ)sin⁡(C−θ)+sin⁡(C+θ)\frac{\cos A - \cos B}{\cos A + \cos B} = \frac{\sin(C - \theta) - \sin(C + \theta)}{\sin(C - \theta) + \sin(C + \theta)}

⇒tan⁡A+B2tan⁡A−B2=cot⁡Ctan⁡θ\tan\frac{A + B}{2}\tan\frac{A - B}{2} = \cot C\tan\theta

⇒ tan⁡θ=tan⁡A+B2tan⁡A−B2tan⁡C\tan\theta = \tan\frac{A + B}{2}\tan\frac{A - B}{2}\tan C