If $\cos{2\alpha = \frac{3\cos{2\beta - 1}}{3 - \cos{2\beta}}}$, then $ an\alpha$ is equal to

Question

If $\cos{2\alpha = \frac{3\cos{2\beta - 1}}{3 - \cos{2\beta}}}$, then $	an\alpha$ is equal to

Dr. Abubacker · Accepted Answer

(a)
Given, $\cos{2\alpha = \frac{2\cos{2\beta - 1}}{3 - \cos{2\beta}}}$
$\Rightarrow \frac{1 - 	an^{2}\alpha}{1 + 	an^{2}\alpha} = \frac{3\left( \frac{1 - 	an^{2}\beta}{1 + 	an^{2}\beta} ight) - 1}{3 - \left( \frac{1 - 	an^{2}\beta}{1 + 	an^{2}\beta} ight)}$
$\Rightarrow \ \frac{1 - 	an^{2}\alpha}{1 + 	an^{2}\alpha} = \frac{2 - 4	an^{2}\beta}{2 + 4	an^{2}\beta} = \frac{1 - 2	an^{2}\beta}{1 + 2	an^{2}\beta}$
Applying componendo and dividendo, we get
$\frac{1}{	an^{2}\alpha} = \frac{1}{2	an^{2}\beta}$
$\Rightarrow 	an{\alpha = \sqrt{2}	an\beta}$

Q. If $\cos{2\alpha = \frac{3\cos{2\beta - 1}}{3 - \cos{2\beta}}}$, then $\tan\alpha$ is equal to

Explanation

Reviewed by Dr. Abubacker

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