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

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(C) We use the identity $	an 	heta = \cot 	heta - 2\cot 2	heta$,which implies $\cot 	heta - 	an 	heta = 2\cot 2	heta$.Consider the expression

Vedclass · Accepted Answer

$\cot \alpha $

Vedclass · Answer

(C) We use the identity $	an 	heta = \cot 	heta - 2\cot 2	heta$,which implies $\cot 	heta - 	an 	heta = 2\cot 2	heta$.Consider the expression $E = 	an \alpha + 2	an 2\alpha + 4	an 4\alpha + 8\cot 8\alpha$.Using the identity $	an 	heta = \cot 	heta - 2\cot 2	heta$,we have:$8\cot 8\alpha + 4	an 4\alpha = 4(2\cot 8\alpha + 	an 4\alpha) = 4(\cot 4\alpha) = 4\cot 4\alpha$.Now the expression becomes:$E = 	an \alpha + 2	an 2\alpha + 4\cot 4\alpha$.Again,using $4\cot 4\alpha + 2	an 2\alpha = 2(2\cot 4\alpha + 	an 2\alpha) = 2(\cot 2\alpha) = 2\cot 2\alpha$.Now the expression becomes:$E = 	an \alpha + 2\cot 2\alpha$.Using $2\cot 2\alpha = \cot \alpha - 	an \alpha$,we get:$E = 	an \alpha + (\cot \alpha - 	an \alpha) = \cot \alpha$.Thus,the correct option is $C$.

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

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