Question

$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha$ is equal to :

• A. $\tan 16\alpha$
• B. 0
• C. $\cot \alpha$
• D. none of these

Answer 1

Answer: (C)

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