Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan \alpha$ 
Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos
 \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan
 \alpha$.

I tried:
$$\frac{\frac{\sin \alpha}{\cos \alpha}+\sin \alpha}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha+\sqrt{5}\sin \alpha \cos \alpha}{\cos \alpha}}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha}{\cos \alpha}(\sqrt{5}\cos \alpha+1)}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\tan \alpha (5\cos^2\alpha-1)}{2} = \\
???$$
What do I do next?

Now simply $$\frac{5\sin\alpha \cos\alpha-\tan\alpha}{2}$$
$$\frac{5\times2.\sin\alpha \cos\alpha}{4}-\frac{\tan\alpha}{2}$$
$$\frac{5\times\sin 2\alpha}{4}-\frac{\tan\alpha}{2}\tag{sin2x=2 sinx cox}$$
