$\tan(\phi-\theta)=\frac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$.
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*}
By the definition of cosine as the x-coordinate of a circle, cosine is negative in the interval $(90^{\circ},180^{\circ})$
$\cos 2x =2\cos^2x-1=1-2\sin^2x$
$\cos 2x =(\sqrt{2} \cos x-1)(\sqrt{2} \cos x+1)=-(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1)$
Using the fact that $\cos 2x$ is -ve in the interval $x \in (45^{\circ},90^{\circ})$
Prove the fact that $\cos x < \frac{1}{\sqrt{2}}< \sin x$ in the interval $x \in (45^{\circ},90^{\circ})$
Proof 2:
There is 1 more way people define $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$
Let the angles be $x,90-x,90$
Use the fact that the side opposite to the greater angle is greater.
Therefore, adjacent < opposite (for $x \in (45^{\circ},90^{\circ})$)
Therefore, $\cos x < \sin x$ for $x \in (45^{\circ},90^{\circ})$
Best Answer
Here’s yet another way:
$$LHS=\frac{1-\cos(\frac{\pi}{2}-2\alpha)}{1+\cos(\frac{\pi}{2}-2\alpha)}$$ $$=\frac{2\sin^2(\frac{\pi}{4}-\alpha)}{2\cos^2(\frac{\pi}{4}-\alpha)}$$ $$=\tan^2(\frac{\pi}{4}-\alpha)$$ $$=\tan^2(\alpha-\frac{\pi}{4}+\pi)$$ $$=RHS$$