Prove: $tan 3 A - tan 2 A - tan A = tan A tan 2 A tan 3 A$ .

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Solution

$L H S = tan 3 A - tan 2 A - tan A = tan (2 A + A) - (tan 2 A + tan A) = \frac{tan 2 A + tan A}{1 - tan 2 A tan A} - (tan 2 A + tan A)$
$= (tan 2 A + tan A) \times \frac{1 - 1 + tan 2 A tan A}{1 - tan 2 A tan A} = \frac{tan 2 A + tan A}{1 - tan 2 A tan A} \times tan 2 A tan A = tan 3 A tan 2 A tan A = R H S$
Hence Proved

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