How prove this $$-\tan\frac{10\pi}{41}+4\left(\sin\frac{2\pi}{41}+\sin\frac{4\pi}{41}+\sin\frac{12\pi}{41}+\sin\frac{20\pi}{41}-\sin\frac{26\pi}{41}-\sin \frac{30\pi}{41}\right)= \\\cot\frac{2\pi}{41}-\cot\frac{8\pi}{41}+\cot\frac{20\pi}{41}+\cot\frac{32\pi}{41}+\cot\frac{36\pi}{41}$$
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4step 1 is to edit your question so that it displays properly. \frac{numerator}{denominator}, \tan, \sin, and \cos would be a good start. – Teepeemm Sep 26 '14 at 20:46
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Perhaps the complex method? – GEdgar Sep 26 '14 at 21:58
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for information,by user the complex method,the common value of these relationships is $\sqrt{41+4\sqrt{41}-2\sqrt{533+80\sqrt{41}}}$ – user178256 Oct 02 '14 at 20:57