Prove that: $\tan (3 x) \tan (2 x) \tan (x) = \tan (3 x) - \tan (2 x) - \tan (x)$

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Solution

Solve for the required proof
Given that $\tan (3 x) \tan (2 x) \tan (x) = \tan (3 x) - \tan (2 x) - \tan (x)$
Consider $\tan (3 x) = \tan (2 x + x)$
The formula to find the tangent of summation of two angles is
$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
Substituting $A = 2 x, B = x$ ,
$\Rightarrow$ $\tan (2 x + x) = \frac{\tan (2 x) + \tan (x)}{1 - \tan (2 x) \tan (x)}$
$\Rightarrow$ $\tan (3 x) (1 - \tan (2 x) \tan (x)) = \tan (2 x) + \tan (x)$
$\Rightarrow$ $\tan (3 x) - \tan (3 x) \tan (2 x) \tan (x) = \tan (2 x) + \tan (x)$
$\Rightarrow$ $\tan (3 x) - \tan (2 x) - \tan (x) = \tan (3 x) \tan (2 x) \tan (x)$
Hence, the given identity is proved.

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