# trig and solve

• May 16th 2012, 09:22 PM
srirahulan
trig and solve
$\cot \frac{\pi}{24}$ = $\frac {2-\sqrt3}{\sqrt6 - \sqrt2 - 1}$Simplify and find out value of the $\cot\frac{\pi}{24}$
• May 17th 2012, 07:38 AM
skeeter
Re: trig and solve
$\frac{2-\sqrt{3}}{\sqrt{6} - (\sqrt{2}+1)} \cdot \frac{\sqrt{6} + (\sqrt{2}+1)}{\sqrt{6} + (\sqrt{2}+1)}$

$\frac{2\sqrt{6} + 2\sqrt{2} + 2 - 3\sqrt{2} - \sqrt{6} - \sqrt{3}}{6 - (2 + 2\sqrt{2} + 1)}$

$\frac{\sqrt{6} - \sqrt{2} - \sqrt{3} + 2}{3 - 2\sqrt{2}} \cdot \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}}$

$\frac{3\sqrt{6} - 3\sqrt{2} - 3\sqrt{3} + 6 + 4\sqrt{3} - 4 - 2\sqrt{6} + 4\sqrt{2}}{9-8}$

$\sqrt{6} + \sqrt{2} + \sqrt{3} + 2$