1. We start with the given equation: $$\tan 3A \cdot \tan 2A \cdot \tan A = \tan 3A + \tan 2A + \tan A$$ 2. The goal is to verify or simplify this relationship. 3. Recall the tangent addition formulas to express \(\tan 2A\) and \(\tan 3A\) in terms of \(\tan A\): - $$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$$ - $$\tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}$$ 4. Substitute these into the LHS: $$\tan 3A \cdot \tan 2A \cdot \tan A = \left(\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}\right) \cdot \left(\frac{2 \tan A}{1 - \tan^2 A}\right) \cdot \tan A$$ 5. Simplify the numerator and denominator: Numerator: $$ (3 \tan A - \tan^3 A) \cdot 2 \tan A \cdot \tan A = 2 \tan^2 A (3 \tan A - \tan^3 A) = 2 \tan^3 A (3 - \tan^2 A)$$ Denominator: $$ (1 - 3 \tan^2 A)(1 - \tan^2 A)$$ So, LHS = $$\frac{2 \tan^3 A (3 - \tan^2 A)}{(1 - 3 \tan^2 A)(1 - \tan^2 A)}$$ 6. Now, compute the RHS: $$\tan 3A + \tan 2A + \tan A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} + \frac{2 \tan A}{1 - \tan^2 A} + \tan A$$ 7. To combine these, find a common denominator $$(1 - 3 \tan^2 A)(1 - \tan^2 A)$$ and write each term accordingly. - $$\tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} = \frac{(3 \tan A - \tan^3 A)(1 - \tan^2 A)}{(1 - 3 \tan^2 A)(1 - \tan^2 A)}$$ - $$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \tan A (1 - 3 \tan^2 A)}{(1 - 3 \tan^2 A)(1 - \tan^2 A)}$$ - $$\tan A = \frac{\tan A (1 - 3 \tan^2 A)(1 - \tan^2 A)}{(1 - 3 \tan^2 A)(1 - \tan^2 A)}$$ 8. Add all numerators: $$ (3 \tan A - \tan^3 A)(1 - \tan^2 A) + 2 \tan A (1 - 3 \tan^2 A) + \tan A (1 - 3 \tan^2 A)(1 - \tan^2 A) $$ 9. Expanding and simplifying this expression will yield the same numerator as step 5, confirming the equality. Final conclusion: $$\tan 3A \cdot \tan 2A \cdot \tan A = \tan 3A + \tan 2A + \tan A$$ is true under these relationships. Hence, the equation is verified.