Question: p tan(\alpha-\beta)\cdot tan(\beta-\alpha)\cdot tan(2\beta-2\alpha) = tan(\beta-\alpha) - tan(2\beta-2\alpha) - tan(\alpha-\beta)
1. **State the problem:**
We want to verify or simplify the expression:
$$p \tan(\alpha-\beta) \cdot \tan(\beta-\alpha) \cdot \tan(2\beta-2\alpha) = \tan(\beta-\alpha) - \tan(2\beta-2\alpha) - \tan(\alpha-\beta)$$
2. **Recall important identities:**
- $\tan(-x) = -\tan(x)$
- $\tan(a) \cdot \tan(-a) = -\tan^2(a)$
3. **Simplify the left side:**
Note that $\tan(\beta-\alpha) = -\tan(\alpha-\beta)$, so
$$\tan(\alpha-\beta) \cdot \tan(\beta-\alpha) = \tan(\alpha-\beta) \cdot (-\tan(\alpha-\beta)) = -\tan^2(\alpha-\beta)$$
Therefore,
$$p \tan(\alpha-\beta) \cdot \tan(\beta-\alpha) \cdot \tan(2\beta-2\alpha) = p \cdot (-\tan^2(\alpha-\beta)) \cdot \tan(2\beta-2\alpha) = -p \tan^2(\alpha-\beta) \tan(2\beta-2\alpha)$$
4. **Rewrite the right side:**
Using $\tan(\beta-\alpha) = -\tan(\alpha-\beta)$,
$$\tan(\beta-\alpha) - \tan(2\beta-2\alpha) - \tan(\alpha-\beta) = -\tan(\alpha-\beta) - \tan(2\beta-2\alpha) - \tan(\alpha-\beta) = -2 \tan(\alpha-\beta) - \tan(2\beta-2\alpha)$$
5. **Conclusion:**
The original equation can be rewritten as
$$-p \tan^2(\alpha-\beta) \tan(2\beta-2\alpha) = -2 \tan(\alpha-\beta) - \tan(2\beta-2\alpha)$$
This relates $p$ to the tangent expressions. To solve for $p$, divide both sides by $-\tan^2(\alpha-\beta) \tan(2\beta-2\alpha)$:
$$p = \frac{-2 \tan(\alpha-\beta) - \tan(2\beta-2\alpha)}{-\tan^2(\alpha-\beta) \tan(2\beta-2\alpha)}$$
6. **Simplify the fraction:**
$$p = \frac{2 \tan(\alpha-\beta) + \tan(2\beta-2\alpha)}{\tan^2(\alpha-\beta) \tan(2\beta-2\alpha)}$$
This is the simplified form for $p$ in terms of tangent functions.
**Final answer:**
$$p = \frac{2 \tan(\alpha-\beta) + \tan(2\beta-2\alpha)}{\tan^2(\alpha-\beta) \tan(2\beta-2\alpha)}$$