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Prove (nested radical)
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\[\frac{\sqrt{10 + \sqrt{1}}+ \sqrt{10 + \sqrt{2}}+ ... + \sqrt{10 + \sqrt{99}}}{\sqrt{10 - \sqrt{1}}+\sqrt{10 - \sqrt{2}}+...+\sqrt{10 - \sqrt{99}}}=\cot \frac{\pi}{8}\]
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.
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