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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.

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  • $\begingroup$ See math.stackexchange.com/questions/472594/… $\endgroup$ – lab bhattacharjee Jan 13 '18 at 17:41
  • $\begingroup$ I have seen this perticular solution... But I was wondering how to apply that in this case $\endgroup$ – Le Connoisseur Jan 13 '18 at 17:49
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Jan 13 '18 at 17:51
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    $\begingroup$ So, $$\cot142.5^\circ=\dfrac{1+\cos285^\circ}{\sin285^\circ}$$ $$285=360-75,75=45+30$$ $\endgroup$ – lab bhattacharjee Jan 13 '18 at 17:51
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use $$\cot(2x)=\frac{1}{2}\frac{\cot(x)-1}{\cot(x)}$$ and $$\cot(285^{\circ})=\sqrt{3}-2$$

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