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

I tried:

$\frac{\frac{\mathrm{sin}\alpha}{\mathrm{cos}\alpha}+\mathrm{sin}\alpha}{2}(\sqrt{5}\mathrm{cos}\alpha -1)=\phantom{\rule{0ex}{0ex}}\frac{\frac{\mathrm{sin}\alpha +\sqrt{5}\mathrm{sin}\alpha \mathrm{cos}\alpha}{\mathrm{cos}\alpha}}{2}(\sqrt{5}\mathrm{cos}\alpha -1)=\phantom{\rule{0ex}{0ex}}\frac{\frac{\mathrm{sin}\alpha}{\mathrm{cos}\alpha}(\sqrt{5}\mathrm{cos}\alpha +1)}{2}(\sqrt{5}\mathrm{cos}\alpha -1)=\phantom{\rule{0ex}{0ex}}\frac{\mathrm{tan}\alpha (5{\mathrm{cos}}^{2}\alpha -1)}{2}=\phantom{\rule{0ex}{0ex}}???$

I tried:

$\frac{\frac{\mathrm{sin}\alpha}{\mathrm{cos}\alpha}+\mathrm{sin}\alpha}{2}(\sqrt{5}\mathrm{cos}\alpha -1)=\phantom{\rule{0ex}{0ex}}\frac{\frac{\mathrm{sin}\alpha +\sqrt{5}\mathrm{sin}\alpha \mathrm{cos}\alpha}{\mathrm{cos}\alpha}}{2}(\sqrt{5}\mathrm{cos}\alpha -1)=\phantom{\rule{0ex}{0ex}}\frac{\frac{\mathrm{sin}\alpha}{\mathrm{cos}\alpha}(\sqrt{5}\mathrm{cos}\alpha +1)}{2}(\sqrt{5}\mathrm{cos}\alpha -1)=\phantom{\rule{0ex}{0ex}}\frac{\mathrm{tan}\alpha (5{\mathrm{cos}}^{2}\alpha -1)}{2}=\phantom{\rule{0ex}{0ex}}???$