tan 3A - tan 2A - tan A = | vedclass

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(A) We know that $3A = 2A + A$.Taking tangent on both sides: $	an 3A = 	an(2A + A)$.Using the formula $	an(x + y) = \frac{	an x + 	an y}{1 - 	an

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$	an 3A 	an 2A 	an A$

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(A) We know that $3A = 2A + A$.Taking tangent on both sides: $	an 3A = 	an(2A + A)$.Using the formula $	an(x + y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we get:$	an 3A = \frac{	an 2A + 	an A}{1 - 	an 2A 	an A}$.Cross-multiplying gives:$	an 3A(1 - 	an 2A 	an A) = 	an 2A + 	an A$.$	an 3A - 	an 3A 	an 2A 	an A = 	an 2A + 	an A$.Rearranging the terms,we get:$	an 3A - 	an 2A - 	an A = 	an 3A 	an 2A 	an A$.

$\tan 3A - \tan 2A - \tan A = $

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