Question

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

• A. 0
• B. $2\tan \alpha$
• C. $\cot \alpha$
• D. $\tan 16$

Answer 1

The correct option is C $\cot \alpha$

Given

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

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

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

$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +4\left(\cot 4\alpha -\tan 4\alpha \right)$

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

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

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

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

$\tan \alpha +2\tan 2\alpha +2\left(\cot 2\alpha -\tan 2\alpha \right)$

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

$\tan \alpha +2\cot 2\alpha$

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

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

$\tan \alpha +\left(\cot \alpha -\tan \alpha \right)$

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

$\cot \alpha$