tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = | vedclass

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(B) We use the identity $\cot 	heta - 	an 	heta = 2 \cot 2	heta$,which implies $\cot 	heta = 	an 	heta + 2 \cot 2	heta$.Rearranging,we get $2

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$\cot A$

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(B) We use the identity $\cot 	heta - 	an 	heta = 2 \cot 2	heta$,which implies $\cot 	heta = 	an 	heta + 2 \cot 2	heta$.Rearranging,we get $2 \cot 2	heta = \cot 	heta - 	an 	heta$.Now,consider the expression $E = 	an A + 2 	an 2A + 4 	an 4A + 8 \cot 8A$.Using the identity $	an 	heta = \cot 	heta - 2 \cot 2	heta$,we have:$	an A = \cot A - 2 \cot 2A$Substituting this into the expression:$E = (\cot A - 2 \cot 2A) + 2 	an 2A + 4 	an 4A + 8 \cot 8A$This approach is complex,so let us simplify step-by-step using $	an 	heta = \cot 	heta - 2 \cot 2	heta$ repeatedly:$8 \cot 8A = 4 \cot 4A - 4 	an 4A$$E = 	an A + 2 	an 2A + 4 	an 4A + (4 \cot 4A - 4 	an 4A) = 	an A + 2 	an 2A + 4 \cot 4A$Now,$4 \cot 4A = 2 \cot 2A - 2 	an 2A$$E = 	an A + 2 	an 2A + (2 \cot 2A - 2 	an 2A) = 	an A + 2 \cot 2A$Finally,$2 \cot 2A = \cot A - 	an A$$E = 	an A + (\cot A - 	an A) = \cot A$.

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

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