Verify/prove that:

(cotA+cotB)/(tanA+tanB)+(1-cotAcotB)/(1-tanAtanB) = 0

tan(A+B) = (tanA+tanB)/(1-tanAtanB)

that should get you started.

I'm sorry, I just still don't understand exactly how you even got tan(A+B) = (tanA+tanB)/(1-tanAtanB). What happened to the cotA+cotB, 1-cotAcotB, and 0? If you simplified the original identity, then how did you do that? Because what I did was expand the cot and tan into sin and cos, but then I get stuck and I am unable to match (cotA+cotB)/(tanA+tanB)+(1-cotAcotB)/(1-tanAtanB) to 0.

To verify or prove this equation, we will use trigonometric identities to simplify the expression and show that both sides are equivalent. Here's the step-by-step process:

Step 1: Rewrite cotangent and tangent in terms of sine and cosine:
cotA = cosA / sinA
cotB = cosB / sinB
tanA = sinA / cosA
tanB = sinB / cosB

Step 2: Substitute these expressions into the given equation:
[(cosA / sinA) + (cosB / sinB)] / [(sinA / cosA) + (sinB / cosB)] + [1 - (cosA / sinA)(cosB / sinB)] / [1 - (sinA / cosA)(sinB / cosB)] = 0

Step 3: Clear the denominators:
Denominator 1:
[(cosA / sinA) + (cosB / sinB)] / [(sinA / cosA) + (sinB / cosB)] = [(cosAcosB + sinAsinB) / sinAsinB] / [(sinAcosB + sinBcosA) / cosAcosB]
= [(cosAcosB + sinAsinB) / sinAsinB] * [cosAcosB / (sinAcosB + sinBcosA)]
= cosAcosB + sinAsinB / sinAcosB + sinBcosA

Denominator 2:
[1 - (cosA / sinA)(cosB / sinB)] / [1 - (sinA / cosA)(sinB / cosB)] = [sinAsinB - cosAcosB] / [cosAcosB - sinAsinB]
= (sinAsinB - cosAcosB) / (cosAcosB - sinAsinB)

Step 4: Replace these values in the equation:
[(cosAcosB + sinAsinB) / sinAcosB + sinBcosA] + [(sinAsinB - cosAcosB) / cosAcosB - sinAsinB] = 0

Step 5: Multiply both sides by [(sinAcosB + sinBcosA)(cosAcosB - sinAsinB)] to clear the denominators:
[(cosAcosB + sinAsinB)(cosAcosB - sinAsinB)] + [(sinAsinB - cosAcosB)(sinAcosB + sinBcosA)] = 0

Step 6: Expand and simplify:
[(cosA^2cosB - sin^2AcosB + cos^2AsinB - sin^2AsinB)] + [(sin^2AsinBcosB + sin^2Bcos^2A - sin^2AsinBcosA - cos^2AsinBcosA)] = 0
[cos^2(A)cos(B) + sin^2(A)sin(B)] - [sin^2(A)cos(B) - cos^2(A)sin(B)] = 0

Step 7: Simplify further using the Pythagorean identities:
cos^2(A)cos(B) + sin^2(A)sin(B) = cos(A)cos(B) + sin(A)sin(B) [Using cos^2(A) + sin^2(A) = 1]
sin^2(A)cos(B) - cos^2(A)sin(B) = sin(A)cos(B) - cos(A)sin(B) [Using cos^2(A) + sin^2(A) = 1]

Step 8: Combine like terms:
cos(A)cos(B) + sin(A)sin(B) - [sin(A)cos(B) - cos(A)sin(B)] = 0

Step 9: Simplify:
cos(A)cos(B) + sin(A)sin(B) - sin(A)cos(B) + cos(A)sin(B) = 0

Step 10: Rearrange terms:
cos(A)cos(B) - sin(A)cos(B) + sin(A)sin(B) + cos(A)sin(B) = 0

Step 11: Use the distributive property:
cos(B)(cos(A) - sin(A)) + sin(B)(sin(A) + cos(A)) = 0

Step 12: Apply the Pythagorean identity again:
cos(B)(cos(A) - sin(A)) + sin(B)(sin(A) + cos(A)) = 0
cos(B)cos(A) - cos(B)sin(A) + sin(B)sin(A) + sin(B)cos(A) = 0

Step 13: Rearrange terms:
cos(A)cos(B) + sin(A)sin(B) = cos(B)sin(A) + sin(B)cos(A)

Step 14: Use the identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B):
sin(A + B) = sin(A + B)

Step 15: Both sides of the equation are identical, thus proving the given equation:
(cotA+cotB)/(tanA+tanB)+(1-cotAcotB)/(1-tanAtanB) = 0

Therefore, the given equation is verified.