What is the sum/difference angle formula for Tan (B-A) ? Quickly please... thank you I can only find Tan (A-B)?

they're the same thing if tan(α − β) = (tan(α) − tan(β))/(1 + tan(α)tan(β))

then tan(B − A) = (tan(B) − tan(A))/(1 + tan(B)tan(A))

just substitute the variables in the same corresponding place

tan(30° − 60°) = (tan(30°) − tan(60°))/(1 + tan(30°)tan(60°))

∴ tan(30° − 60°) = (√(3)/3 − √(3))/(1 + (√(3)/3)(√3))

∴ tan(30° − 60°) = (√(3)/3 − 3√(3)/3)/(1 + 3/3)

∴ tan(30° − 60°) = (-2√(3)/3)/2

∴ tan(30° − 60°) = -√(3)/3

tan(60° − 30°) = (tan(60°) − tan(30°))/(1 + tan(60°)tan(30°))

∴ tan(60° − 30°) = (√(3) − √(3)/3)/(1 + √(3)(√(3)/3))

∴ tan(60° − 30°) = (3√(3)/3 − √(3)/3)/(1 + 3/3)

∴ tan(60° − 30°) = (2√(3)/3)/2

∴ tan(60° − 30°) = √(3)/3

Edit however if it will help you understand

tan(B − A) = tan(-(A − B))

given tan(-θ) = -tan(θ)

∴ tan(B − A) = -tan(A − B)

Duuuuuhhhhh!

Use the Tan (A-B) formula, but just switch around the A and the B !!!!!!!!