# Show that tan &#x2061;<!-- ⁡ --> &#x03B1;<!-- α --> + <msqrt>

Show that $\frac{\mathrm{tan}\alpha +\sqrt{5}\mathrm{sin}\alpha }{2}\left(\sqrt{5}\mathrm{cos}\alpha -1\right)$ is equivalent to $\frac{5}{4}\mathrm{sin}\left(2\alpha \right)-\frac{1}{2}\mathrm{tan}\alpha$
I tried:
$\frac{\frac{\mathrm{sin}\alpha }{\mathrm{cos}\alpha }+\mathrm{sin}\alpha }{2}\left(\sqrt{5}\mathrm{cos}\alpha -1\right)=\phantom{\rule{0ex}{0ex}}\frac{\frac{\mathrm{sin}\alpha +\sqrt{5}\mathrm{sin}\alpha \mathrm{cos}\alpha }{\mathrm{cos}\alpha }}{2}\left(\sqrt{5}\mathrm{cos}\alpha -1\right)=\phantom{\rule{0ex}{0ex}}\frac{\frac{\mathrm{sin}\alpha }{\mathrm{cos}\alpha }\left(\sqrt{5}\mathrm{cos}\alpha +1\right)}{2}\left(\sqrt{5}\mathrm{cos}\alpha -1\right)=\phantom{\rule{0ex}{0ex}}\frac{\mathrm{tan}\alpha \left(5{\mathrm{cos}}^{2}\alpha -1\right)}{2}=\phantom{\rule{0ex}{0ex}}???$
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odmeravan5c
Now simply
$\frac{5\mathrm{sin}\alpha \mathrm{cos}\alpha -\mathrm{tan}\alpha }{2}$
$\frac{5×2.\mathrm{sin}\alpha \mathrm{cos}\alpha }{4}-\frac{\mathrm{tan}\alpha }{2}$
$\begin{array}{}\text{(sin2x=2 sinx cox)}& \frac{5×\mathrm{sin}2\alpha }{4}-\frac{\mathrm{tan}\alpha }{2}\end{array}$