tan alpha + 2 tan 2 alpha + 4 tan 4 alpha + 8 | vedclass

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(C) We use the identity $	an 	heta - \cot 	heta = -2 \cot 2 	heta$. Starting with the expression: $S = 	an \alpha + 2 	an 2 \alpha + 4 	an 4 \a

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$\cot \alpha$

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(C) We use the identity $	an 	heta - \cot 	heta = -2 \cot 2 	heta$. Starting with the expression: $S = 	an \alpha + 2 	an 2 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$. Using $	an \alpha = \cot \alpha - 2 \cot 2 \alpha$,we substitute: $S = (\cot \alpha - 2 \cot 2 \alpha) + 2 	an 2 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha + 2(	an 2 \alpha - \cot 2 \alpha) + 4 	an 4 \alpha + 8 \cot 8 \alpha$ Since $	an 2 \alpha - \cot 2 \alpha = -2 \cot 4 \alpha$: $S = \cot \alpha + 2(-2 \cot 4 \alpha) + 4 	an 4 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha - 4 \cot 4 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha + 4(	an 4 \alpha - \cot 4 \alpha) + 8 \cot 8 \alpha$ Since $	an 4 \alpha - \cot 4 \alpha = -2 \cot 8 \alpha$: $S = \cot \alpha + 4(-2 \cot 8 \alpha) + 8 \cot 8 \alpha$ $S = \cot \alpha - 8 \cot 8 \alpha + 8 \cot 8 \alpha$ $S = \cot \alpha$.

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

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