Answer 1

\[\tan A \tan B \tan C = \tan A \tan B \tan(A + B) \left[ \text{ Using } A + B = C \right]\]
\[ = \tan A \tan B \times \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
\[ = \frac{\tan^2 A\tan B + \tan A \tan^2 B}{1 - \tan A \tan B}\]
\[ = \frac{\tan^2 A\tan B + \tan A \tan^2 B + \tan A + \tan B - \tan A - \tan B}{1 - \tan A \tan B}\]
\[ = \frac{- \tan A(1 - \tan A\tan B) - \tan B(1 - \tan A\tan B) + \tan A + \tan B}{1 - \tan A \tan B}\]
\[ = \frac{- (1 - \tan A\tan B)\left( \tan A + \tan B \right) + \tan A + \tan B}{1 - \tan A \tan B}\]
\[ = \frac{\tan A + \tan B}{1 - \tan A \tan B} - \tan A - \tan B \]
\[ = \tan(A + B) - \tan A - \tan B\]
\[ = \tan C - \tan A - \tan B\]
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