Question: If $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$, then show that $x + y + z = xyz$
1. **State the problem:**
We are given that $$\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$$ and we need to prove that $$x + y + z = xyz$$.
2. **Recall the formula for the sum of two inverse tangents:**
$$\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right)$$ provided the result is in the correct quadrant.
3. **Apply the formula to the first two terms:**
Let $$A = \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right)$$.
4. **Rewrite the original equation:**
$$A + \tan^{-1} z = \pi$$
which implies
$$\tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z = \pi$$.
5. **Use the tangent addition formula again:**
$$\tan \left( \tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z \right) = \tan \pi = 0$$.
6. **Express the tangent of the sum:**
$$\tan \left( \tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z \right) = \frac{ \frac{x + y}{1 - xy} + z }{ 1 - \left( \frac{x + y}{1 - xy} \right) z } = 0$$.
7. **Set the numerator equal to zero:**
$$\frac{x + y}{1 - xy} + z = 0$$
8. **Multiply both sides by $1 - xy$:**
$$x + y + z(1 - xy) = 0$$
9. **Expand:**
$$x + y + z - xyz = 0$$
10. **Rearrange to isolate terms:**
$$x + y + z = xyz$$
**Final answer:**
$$x + y + z = xyz$$
This completes the proof.