# If $\tan\theta =\cos2\alpha\tan\phi$ then prove that $\tan(\phi-\theta)=\frac{\tan^2\alpha\sin2\phi}{1+\tan^2\alpha\cos2\phi}$

If $$\tan\theta =\cos2\alpha\tan\phi$$ then prove that $$\tan(\phi-\theta)=\frac{\tan^2\alpha\sin2\phi}{1+\tan^2\alpha\cos2\phi}$$

I have tried in a few different ways applying the formula $$\tan(\phi - \theta) = \frac{\tan \phi - \tan \theta} {1+\tan \phi \tan \theta}$$ then substituting $$\tan \theta = \cos 2\alpha \tan \phi$$. But all in vain. Assuming the problem to be correct, my highest achievement was getting the numerator correct, but the denominator really gives me a lot of headache. Please guide me. Thanks in advance.

$$\tan(\phi-\theta)=\frac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$$.
If we replace $$\tan\theta=\cos 2\alpha \tan \phi$$ we get
\begin{align*} \frac{\tan\phi-\cos 2\alpha\tan\phi}{1+\tan\phi\cos2\alpha\tan\phi}&=\frac{\tan\phi-(\cos^2\alpha-\sin^2 \alpha)\tan\phi}{1+\tan^2\phi(\cos^2\alpha-\sin^2\alpha)}\\ &=\frac{2\sin^2\alpha\tan\phi\cos^2\phi}{\cos^2\phi+\sin^2\phi(\cos^2\alpha-\sin^2\alpha)}\\ &=\frac{2\sin^2\alpha \sin\phi\cos\phi}{\cos^2\phi+\sin^2\phi(1-2\sin^2\alpha)}\\ &=\frac{\sin^2\alpha \sin 2\phi}{1-2\sin^2\alpha\sin^2\phi}\\ &=\frac{\sin^2\alpha \sin 2\phi}{\cos^2\alpha+\sin^2\alpha-2\sin^2\alpha\sin^2\phi}\\ &=\frac{\sin^2\alpha\sin 2\phi}{\cos^2\alpha+\sin^2\alpha(1-2\sin^2\phi)}\\&=\frac{\sin^2\alpha\sin2\phi}{\cos^2\alpha(1+\tan^2\alpha \cos2\phi)}\\ &=\frac{\tan^2\alpha \sin 2\phi }{1+\tan^2\alpha \cos2\phi}. \end{align*}