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This is the problem: Prove that $\tan(A) + \tan(B) + \tan(C) = \tan(A) \tan(B) \tan(C)$.
The problem above assumes that triangle ABC is not a right triangle, as $\tan(90)$ is undefined.
First we know that angles add up to 180.
$$A + B + C = 180$$
$$A + B = 180 - C$$
$$\tan(A + B) = \tan(180-C)$$
Now, we can use two identities:
$$\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$$
$$\tan(180 - C) = -\tan(C)$$
The bottom identity is true because 180 - $\theta$ is essentially the same angle, but in the other quadrant, so the cosine of the angle is negative, but sin stays the same, so tan is negated. Now, we set them equal to each other.
$$\frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} = -\tan(C)$$
Now, we multiply by the denominator on both sides.
$$\tan(A) + \tan(B) = -\tan(C)\cdot(1 - \tan(A)\tan(B))$$
$$\tan(A) + \tan(B) = -\tan(C) + \tan(A)\tan(B)\tan(C)$$
$$\boxed{\tan(A) + \tan(B) + \tan(C) = \tan(A)\tan(B)\tan(C)}$$
Now, we have proved that the sum of the tangents of the angles in a triangle equals their product.