Prove that tan 4x = frac{4 tan x (1 - tan^2 x | vedclass

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We know that $	an 2A = \frac{2 	an A}{1 - 	an^2 A}$.$L.H.S. = 	an 4x = 	an 2(2x)$.Using the formula $	an 2A = \frac{2 	an A}{1 - 	an^2 A}$ wit

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We know that $	an 2A = \frac{2 	an A}{1 - 	an^2 A}$.$L.H.S. = 	an 4x = 	an 2(2x)$.Using the formula $	an 2A = \frac{2 	an A}{1 - 	an^2 A}$ with $A = 2x$:$= \frac{2 	an 2x}{1 - 	an^2 2x}$.Substituting $	an 2x = \frac{2 	an x}{1 - 	an^2 x}$:$= \frac{2 \left( \frac{2 	an x}{1 - 	an^2 x} ight)}{1 - \left( \frac{2 	an x}{1 - 	an^2 x} ight)^2}$.$= \frac{\frac{4 	an x}{1 - 	an^2 x}}{1 - \frac{4 	an^2 x}{(1 - 	an^2 x)^2}}$.$= \frac{\frac{4 	an x}{1 - 	an^2 x}}{\frac{(1 - 	an^2 x)^2 - 4 	an^2 x}{(1 - 	an^2 x)^2}}$.$= \frac{4 	an x (1 - 	an^2 x)}{(1 - 	an^2 x)^2 - 4 	an^2 x}$.$= \frac{4 	an x (1 - 	an^2 x)}{1 + 	an^4 x - 2 	an^2 x - 4 	an^2 x}$.$= \frac{4 	an x (1 - 	an^2 x)}{1 - 6 	an^2 x + 	an^4 x} = R.H.S.$

Prove that $\tan 4x = \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x}$.

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