If \( \cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta} \), Find \( \tan\alpha \)
Question
If \(\alpha\) and \(\beta\) are acute angles satisfying
\[ \cos2\alpha = \frac{3\cos2\beta-1} {3-\cos2\beta}, \]
then \(\tan\alpha\) is equal to
(a) \(\sqrt2\tan\beta\)
(b) \(\frac1{\sqrt2}\tan\beta\)
(c) \(\sqrt2\cot\beta\)
(d) \(\frac1{\sqrt2}\cot\beta\)
Solution
Use the identity
\[ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta} \]
Let
\[ t=\tan\beta \]
Then
\[ \cos2\beta = \frac{1-t^2}{1+t^2} \]
Substitute into the given expression:
\[ \cos2\alpha = \frac{3\left(\frac{1-t^2}{1+t^2}\right)-1} {3-\left(\frac{1-t^2}{1+t^2}\right)} \]
\[ = \frac{\frac{3-3t^2-(1+t^2)}{1+t^2}} {\frac{3(1+t^2)-(1-t^2)}{1+t^2}} \]
\[ = \frac{2-4t^2}{2+4t^2} \]
\[ = \frac{1-2t^2}{1+2t^2} \]
Comparing with
\[ \cos2\alpha = \frac{1-\tan^2\alpha} {1+\tan^2\alpha}, \]
we obtain
\[ \tan^2\alpha = 2t^2 = 2\tan^2\beta \]
Since \(\alpha\) and \(\beta\) are acute angles,
\[ \tan\alpha = \sqrt2\,\tan\beta \]
Final Answer
\[ \boxed{\tan\alpha=\sqrt2\,\tan\beta} \]
Hence, the correct option is (a) \(\sqrt2\tan\beta\).