Question

Prove (nested radical)

Answer 1

\[\frac{\sqrt{10 + \sqrt{1}}+ \sqrt{10 + \sqrt{2}}+ ... + \sqrt{10 + \sqrt{99}}}{\sqrt{10 - \sqrt{1}}+\sqrt{10 - \sqrt{2}}+...+\sqrt{10 - \sqrt{99}}}=\cot \frac{\pi}{8}\]

Answer 2

this might help you:
tan(pi/2^n) =\[\frac{ \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}} }{ \sqrt{2+\sqrt{2+...+\sqrt{2}}} }\] , where there are n−1 twos under the nested radical signs.