[Math] If $\cos\alpha = \frac{2\cos\beta – 1}{2-\cos\beta}$ , $(0<\alpha , \beta< \pi)$, then $\tan\frac{\alpha}{2}\cot\frac{\beta}{2}$ is equal to

trigonometry

If $\cos\alpha = \frac{2\cos\beta – 1}{2-\cos\beta}$ , $(0<\alpha , \beta< \pi)$, then $\tan\frac{\alpha}{2}\cot\frac{\beta}{2}$ is equal to?

MY ATTEMPT:
I tried simplifying the equation to get a relation between $\cos\alpha$ and $\cos\beta$. But $\tan$ and $\cot$ aren't coming.
Please, help.

Best Answer

HINT:

Use $\cos2x=\dfrac{1-\tan^2x}{1+\tan^2x}$ for $\cos\alpha,\cos\beta$

Then apply Componendo and Dividendo