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Question: If \(\cos\alpha = \frac{2\cos\beta - 1}{2 - \cos\beta}(0 < \alpha,\beta < \pi)\), then \(\frac{\tan\...

If cos⁡α=2cos⁡β−12−cos⁡β(0<α,β<π)\cos\alpha = \frac{2\cos\beta - 1}{2 - \cos\beta}(0 < \alpha,\beta < \pi), then tan⁡α2tan⁡β2\frac{\tan\frac{\alpha}{2}}{\tan\frac{\beta}{2}} is equal to

A

1

B

2\sqrt{2}

C

3\sqrt{3}

D

13\frac{1}{\sqrt{3}}

Answer

3\sqrt{3}

Explanation

Solution

From the given relation we have

1+cos⁡α\mathbf{1 +}\mathbf{\cos}\mathbf{\alpha} = 1+2cos⁡β−12−cos⁡β=2−cos⁡β+2cos⁡β−12−cos⁡β\mathbf{1 +}\frac{\mathbf{2}\mathbf{\cos}\mathbf{\beta}\mathbf{-}\mathbf{1}}{\mathbf{2}\mathbf{-}\mathbf{\cos}\mathbf{\beta}}\mathbf{=}\frac{\mathbf{2}\mathbf{-}\mathbf{\cos}\mathbf{\beta}\mathbf{+ 2}\mathbf{\cos}\mathbf{\beta}\mathbf{-}\mathbf{1}}{\mathbf{2}\mathbf{-}\mathbf{\cos}\mathbf{\beta}}

or 2cos⁡2α2=1+cos⁡β2−cos⁡β=2cos⁡2(β2)1+2sin⁡2(β2)2\cos^{2}\frac{\alpha}{2} = \frac{1 + \cos\beta}{2 - \cos\beta} = \frac{2\cos^{2}\left( \frac{\beta}{2} \right)}{1 + 2\sin^{2}\left( \frac{\beta}{2} \right)}

or cos⁡2(α2)=cos⁡2(β2)1+2sin⁡2(β2)\cos^{2}\left( \frac{\alpha}{2} \right) = \frac{\cos^{2}\left( \frac{\beta}{2} \right)}{1 + 2\sin^{2}\left( \frac{\beta}{2} \right)} ….……. (1)

⇒ 1−cos⁡2α2=1−cos⁡2(β2)1+2sin⁡2(β2)=1+2sin⁡2(β2)−cos⁡2(β2)1+2sin⁡2(β2)1 - \cos^{2}\frac{\alpha}{2} = 1 - \frac{\cos^{2}\left( \frac{\beta}{2} \right)}{1 + 2\sin^{2}\left( \frac{\beta}{2} \right)} = \frac{1 + 2\sin^{2}\left( \frac{\beta}{2} \right) - \cos^{2}\left( \frac{\beta}{2} \right)}{1 + 2\sin^{2}\left( \frac{\beta}{2} \right)}

or sin⁡2α2=3sin⁡2(β2)1+2sin⁡2(β2)\sin^{2}\frac{\alpha}{2} = \frac{3\sin^{2}\left( \frac{\beta}{2} \right)}{1 + 2\sin^{2}\left( \frac{\beta}{2} \right)} …..……. (2)

From (1) and (2) we get

tan⁡2α2=3tan⁡2β2\tan^{2}\frac{\alpha}{2} = 3\tan^{2}\frac{\beta}{2}⇒tan⁡(α2)tan⁡(β2)=3\frac{\tan\left( \frac{\alpha}{2} \right)}{\tan\left( \frac{\beta}{2} \right)} = \sqrt{3}