Determine all triangles ABC for which tan(A-B)+tan(B-C)+tan(C-A)=0.

There's a hint: "Can you relate A-B to B-C and C-A?"

Should I apply the tangent difference formula (tan(x-y))? Help would be appreciated, thanks.

You will find a nice discussion at

http://answers.tutorvista.com/1101359/trigonometry-application-rule.html

showing that if A+B+C=180,

tan(A-B)+tan(B-C)+tan(C-A) = tan(A-B)*tan(B-C)*tan(C-A)

Given that, one of the factors must be zero. So, A=B or B=C or A=C. That is, the triangle is isosceles.

Thanks steve!

Yes, applying the tangent difference formula can be helpful in solving this problem. The tangent difference formula states that tan(x - y) = (tan(x) - tan(y)) / (1 + tan(x)tan(y)).

Let's start by rewriting the given equation using the tangent difference formula:

tan(A - B) + tan(B - C) + tan(C - A) = 0

Using the tangent difference formula for each term, we have:

[(tan(A) - tan(B))/(1 + tan(A)tan(B))] + [(tan(B) - tan(C))/(1 + tan(B)tan(C))] + [(tan(C) - tan(A))/(1 + tan(C)tan(A))] = 0

Now, let's simplify the equation. To do this, we need to find a common denominator for the three fractions. The common denominator is (1 + tan(A)tan(B))(1 + tan(B)tan(C))(1 + tan(C)tan(A)). Multiply each term by this common denominator:

[(tan(A) - tan(B))(1 + tan(B)tan(C))(1 + tan(C)tan(A))] + [(tan(B) - tan(C))(1 + tan(C)tan(A))(1 + tan(A)tan(B))] + [(tan(C) - tan(A))(1 + tan(A)tan(B))(1 + tan(B)tan(C))] = 0

Expanding and simplifying:

(tan(A) - tan(B))(1 + tan(B)tan(C))(1 + tan(C)tan(A)) + (tan(B) - tan(C))(1 + tan(C)tan(A))(1 + tan(A)tan(B)) + (tan(C) - tan(A))(1 + tan(A)tan(B))(1 + tan(B)tan(C)) = 0

Now, let's rewrite in terms of the given variables:

[(tan(A) - tan(B))(1 + tan(B)tan(C))(1 + tan(C)tan(A))] + [(tan(B) - tan(C))(1 + tan(C)tan(A))(1 + tan(A)tan(B))] + [(tan(C) - tan(A))(1 + tan(A)tan(B))(1 + tan(B)tan(C))] = 0

After expanding and simplifying, we can see that most terms will cancel out when we multiply everything together. Since tan(x) - tan(y) = (tan(x) - tan(y))/(1 + tan(x)tan(y)), the numerator will cancel out with the denominator in each term, except for the terms involving tan(B)tan(C) and tan(A)tan(B)tan(C). So, we are left with:

(tan(B)tan(C) - tan(A)tan(B)tan(C)) + (tan(C)tan(A) - tan(B)tan(C)tan(A)) + (tan(A)tan(B) - tan(A)tan(B)tan(C)) = 0

At this point, we can observe that the terms involving tan(B)tan(C), tan(C)tan(A), and tan(A)tan(B) have a common factor of tan(A)tan(B)tan(C). Factoring out this common factor, we have:

tan(B)tan(C) + tan(C)tan(A) + tan(A)tan(B) - tan(A)tan(B)tan(C) = 0

Finally, we can simplify this expression to:

tan(A)tan(B) + tan(B)tan(C) + tan(C)tan(A) - tan(A)tan(B)tan(C) = 0

Now we have the simplified equation. To find the values of A, B, and C that satisfy this equation, you can substitute different values for the variables and solve for the remaining unknowns.