Question

The value of $\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha -\cot \alpha$.

Answer 1

$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha -\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +\frac{8}{\tan 8\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +\frac{8}{\tan 2\times 4\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +\frac{8}{\tan 8\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +\frac{8(1-{\tan }^{2}4\alpha )}{2\tan 4\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +\frac{4(1-{\tan }^{2}4\alpha )}{\tan 4\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +\frac{8(1-{tan}^{2}4\alpha )}{2tan4\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +\frac{4{\tan }^{2}4\alpha +4-4{\tan }^{2}4\alpha }{\tan 4\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +\frac{4}{\tan 4\alpha }-\cot \alpha$

$=\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +\frac{2(1-{\tan }^{2}2\alpha )}{\tan 2\alpha }-\cot \alpha$

$=\tan \alpha +\frac{2{\tan }^{2}2\alpha +2-2{\tan }^{2}2\alpha }{\tan 2\alpha }-\cot \alpha$

$=\tan \alpha +\frac{2}{\tan 2\alpha }-\cot \alpha$

$=\tan \alpha +\frac{2(1-{\tan }^2\alpha )}{2\tan \alpha }-\cot \alpha$

$=\tan \alpha +\frac{1-{\tan }^2\alpha }{\tan \alpha }-\cot \alpha$

$=\tan \alpha +\cot \alpha -\tan \alpha -\cot \alpha$

$=0$