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 :)
Good start. You got $$\cot \theta= \sqrt{\frac{1+\frac{\sqrt3-1}{2\sqrt2}}{1-\frac{\sqrt3-1}{2\sqrt2}}}=\sqrt{\frac{2\sqrt2+\sqrt3-1}{2\sqrt2-(\sqrt3-1)}}\; .$$
To simplify, multiply numerator and denominator by $2\sqrt2+\sqrt3-1.$
Note that the denominator then becomes $8-(\sqrt3-1)^2=(\sqrt3+1)^2. $
Thus $$\cot \theta= -\frac{2\sqrt2+\sqrt3-1}{\sqrt3+1}.$$
Finally, note that $(\sqrt2+\sqrt3)(\sqrt2-1)(\sqrt3+1)=2\sqrt2+\sqrt3-1,$ so
$\cot \theta=-(\sqrt2+\sqrt3)(\sqrt2-1),$ as desired.
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.
