# Finding the value of cot 142.5°

I have seen a few solutions but I can't apply them to this particular question. The question is:

Prove that $\cot 142.5° = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6}$

Help would be really appreciated.

• – lab bhattacharjee Jan 13 '18 at 17:41
• I have seen this perticular solution... But I was wondering how to apply that in this case – Le Connoisseur Jan 13 '18 at 17:49
• @LeConnoisseur: so you are basically asking what is the relation between $\cot 7.5^\circ$ and $\cot (150-7.5)^\circ$, which should be simple to grasp. – Jack D'Aurizio Jan 13 '18 at 17:51
• So, $$\cot142.5^\circ=\dfrac{1+\cos285^\circ}{\sin285^\circ}$$ $$285=360-75,75=45+30$$ – lab bhattacharjee Jan 13 '18 at 17:51

use $$\cot(2x)=\frac{1}{2}\frac{\cot(x)-1}{\cot(x)}$$ and $$\cot(285^{\circ})=\sqrt{3}-2$$