Question

Prove that:
$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha =\cot \alpha$

Answer 1

$tan\alpha +2tan2\alpha +4tan\alpha +8cot8\alpha =cot\alpha$

Let $tan\theta +2cot2\theta$

$=\frac{sin\theta }{cos\theta }+\frac{2cos2\theta }{sin2\theta }$

$=\frac{sin\theta }{cos\theta }+\frac{2(1-2si{n}^2\theta )}{2sin\theta cos\theta }$

$=\frac{si{n}^2\theta +1-2si{n}^2\theta }{sin\theta cos\theta }$

$=\frac{1-si{n}^2\theta }{sin\theta cos\theta }=\frac{co{s}^2\theta }{sin\theta cos\theta }=\frac{cos\theta }{sin\theta }=cot\theta$

$=tan\alpha +2tan2\alpha +4tan4\alpha +8cot8\alpha$

$=tan\alpha +2tan2\alpha +4(tan4\alpha +2cot2(4\alpha \left)\right)$

$=tan\alpha +2tan2\alpha +4cot4\alpha$

$=tan\alpha +2(tan2\alpha +2cot2(2\alpha \left)\right)$

$=tan\alpha +2cot2\alpha$

$=cot\alpha =RHS.$