Browse Questions

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

Toolbox:
• $\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\tan\alpha+2\tan 2\alpha+4\tan 4\alpha+8\cot 8\alpha=\cot\alpha$
(Or) $\cot\alpha-\tan\alpha-2\tan 2\alpha-4\tan 4\alpha-8\cot 8\alpha=0$
$\cot\alpha-\tan \alpha=\large\frac{1}{\tan \alpha}-$$\tan\alpha \Rightarrow \large\frac{1-\tan^2\alpha}{\tan \alpha} \Rightarrow \large\frac{2}{\tan 2\alpha} \Rightarrow 2\cot 2\alpha LHS 2\cot 2\alpha-2\tan 2\alpha-4\tan 4\alpha-8\cot 8\alpha \Rightarrow 2(\large\frac{1}{\tan 2\alpha}-$$\tan 2\alpha)-4\tan 4\alpha-8\cot 8\alpha$
$\Rightarrow 2\large\frac{1-\tan^2 2\alpha}{\tan 2\alpha}$$-4\tan 4\alpha-8\cot 8\alpha \Rightarrow \large\frac{4(1-\tan^2 2\alpha)}{2\tan 2\alpha}$$-4\tan 4\alpha-8\cot 8\alpha$
$\Rightarrow \large\frac{4}{\tan 4\alpha}$$-4\tan 4\alpha-8\cot 8\alpha \Rightarrow \large\frac{4(1-\tan^24\alpha)}{\tan 4\alpha}$$-8\cot 8\alpha$
$\Rightarrow \large\frac{8(1-\tan^24\alpha)}{2\tan 4\alpha}$$-8\cot 8\alpha \Rightarrow \large\frac{8}{\tan 8\alpha}$$-8\cot 8\alpha$
$\Rightarrow 8\cot 8\alpha-8\cot 8\alpha$=0=RHS
Hence proved