tan 2 α · tan 30 ° - α + tan 2 α · tan 60 ° - α + tan 60 ° -…

Given that, $\tan 2\alpha ·\tan \left(30°-\alpha \right)+\tan 2\alpha ·\tan \left(60°-\alpha \right)+\tan \left(60°-\alpha \right)·\tan \left(30°-\alpha \right)$

$\Rightarrow \tan 2\alpha \left[\tan \left(30°-\alpha \right)+\tan \left(60°-\alpha \right)\right]+\tan \left(60°-\alpha \right)·\tan \left(30°-\alpha \right)$

Using $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$, $\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ we get,

$\tan \left(30°-\alpha \right)=\frac{1-\sqrt{3}\tan \alpha }{\sqrt{3}+\tan \alpha }$ and $\tan \left(60°-\alpha \right)=\frac{\sqrt{3}-\tan \alpha }{1+\sqrt{3}\tan \alpha }$

Putting these values we get,

$\frac{2\tan \alpha }{1-{\tan }^2\alpha }\left[\frac{1-\sqrt{3}\tan \alpha }{\sqrt{3}+\tan \alpha }+\frac{\sqrt{3}-\tan \alpha }{1+\sqrt{3}\tan \alpha }\right]+\left(\frac{1-\sqrt{3}\tan \alpha }{\sqrt{3}+\tan \alpha }\right)·\left(\frac{\sqrt{3}-\tan \alpha }{1+\sqrt{3}\tan \alpha }\right)$

$=\frac{2\tan \alpha }{1-{\tan }^2\alpha }\left[\frac{1-3{\tan }^2\alpha +3-{\tan }^2\alpha }{\left(\sqrt{3}+\tan \alpha \right)\left(1+\sqrt{3}\tan \alpha \right)}\right]+\left[\frac{\sqrt{3}-3\tan \alpha -\tan \alpha +\sqrt{3}{\tan }^2\alpha }{\left(\sqrt{3}+\tan \alpha \right)\left(1+\sqrt{3}\tan \alpha \right)}\right]$

$=\frac{1}{\left(\sqrt{3}+\tan \alpha \right)\left(1+\sqrt{3}\tan \alpha \right)}\left[\frac{8\left(1-{\tan }^2\alpha \right)\tan \alpha }{\left(1-{\tan }^2\alpha \right)}+\sqrt{3}\left(1+{\tan }^2\alpha \right)-4\tan \alpha \right]$

$=\frac{4\tan \alpha +\sqrt{3}\left(1+{\tan }^2\alpha \right)}{\sqrt{3}+3\tan \alpha +\tan \alpha +\sqrt{3}{\tan }^2\alpha }$

$=\frac{4\tan \alpha +\sqrt{3}\left(1+{\tan }^2\alpha \right)}{4\tan \alpha +\sqrt{3}\left(1+{\tan }^2\alpha \right)}$

$=1$