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

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(C) We use the identity $\cot 	heta - 	an 	heta = 2\cot 2	heta$,which implies $	an 	heta = \cot 	heta - 2\cot 2	heta$.Applying this to the las

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$\cot \alpha$

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(C) We use the identity $\cot 	heta - 	an 	heta = 2\cot 2	heta$,which implies $	an 	heta = \cot 	heta - 2\cot 2	heta$.Applying this to the last term: $8\cot 8\alpha = 4(2\cot 8\alpha) = 4(\cot 4\alpha - 	an 4\alpha)$.Substituting this into the expression:$= 	an \alpha + 2	an 2\alpha + 4	an 4\alpha + 4(\cot 4\alpha - 	an 4\alpha)$$= 	an \alpha + 2	an 2\alpha + 4\cot 4\alpha$Now,apply the identity again for $4\cot 4\alpha = 2(2\cot 4\alpha) = 2(\cot 2\alpha - 	an 2\alpha)$:$= 	an \alpha + 2	an 2\alpha + 2(\cot 2\alpha - 	an 2\alpha)$$= 	an \alpha + 2\cot 2\alpha$Finally,apply the identity for $2\cot 2\alpha = \cot \alpha - 	an \alpha$:$= 	an \alpha + (\cot \alpha - 	an \alpha)$$= \cot \alpha$.

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

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