Show that $\cot{142\frac{1}{2}^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$. 
Show that $\cot{142\frac{1}{2} ^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.

What I have tried:
Let $\theta = 142\frac{1}{2}^\circ \text{ and }  2\theta = 285^\circ$.
$$\cos 285^\circ = \cos 75^\circ$$
$$\cos 75^\circ = \frac{\sqrt3 - 1}{2\sqrt2}$$
$$\cot \theta = \sqrt{\frac{1 + \cos 2\theta}{1 - \cos 2\theta}}$$
From here, pls help me proceed further. Thank you :)

First render $\cot(142\frac{1}{2}°)=-\cot(37\frac{1}{2}°)$.  Then for the latter angle put in:
$\cot x= \dfrac{2\cos^2 x}{2\sin x \cos x}=\dfrac{1+\cos 2x}{\sin2x}$
with $x=37\frac{1}{2}°$ so $2x=75°$.  If you know the sine and cosine of the latter the rest is algebra.
