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

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(B) We use the identity $\cot 	heta = \frac{1 - 	an^2(	heta/2)}{2 	an(	heta/2)}$ or $\cot 	heta = \frac{1}{	an 	heta}$.Starting with the expre

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$\cot A$

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(B) We use the identity $\cot 	heta = \frac{1 - 	an^2(	heta/2)}{2 	an(	heta/2)}$ or $\cot 	heta = \frac{1}{	an 	heta}$.Starting with the expression: $	an A + 2 	an 2A + 4 	an 4A + 8 \cot 8A$$= 	an A + 2 	an 2A + 4 	an 4A + 8 \left( \frac{1 - 	an^2 4A}{2 	an 4A} ight)$$= 	an A + 2 	an 2A + 4 	an 4A + \frac{4(1 - 	an^2 4A)}{	an 4A}$$= 	an A + 2 	an 2A + \frac{4 	an^2 4A + 4 - 4 	an^2 4A}{	an 4A}$$= 	an A + 2 	an 2A + \frac{4}{	an 4A} = 	an A + 2 	an 2A + 4 \cot 4A$$= 	an A + 2 	an 2A + 4 \left( \frac{1 - 	an^2 2A}{2 	an 2A} ight)$$= 	an A + 2 	an 2A + \frac{2(1 - 	an^2 2A)}{	an 2A}$$= 	an A + \frac{2 	an^2 2A + 2 - 2 	an^2 2A}{	an 2A} = 	an A + \frac{2}{	an 2A}$$= 	an A + 2 \cot 2A = 	an A + 2 \left( \frac{1 - 	an^2 A}{2 	an A} ight)$$= 	an A + \frac{1 - 	an^2 A}{	an A} = \frac{	an^2 A + 1 - 	an^2 A}{	an A} = \frac{1}{	an A} = \cot A$

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

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