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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.