The addition formula tan^(-1)(x) + tan^(-1)(y) has three cases based on the sign of xy: Case 1 (xy < 1): tan^(-1)(x) + tan^(-1)(y) = tan^(-1)$\frac{(x+y}{(1-xy)}$). Case 2 (xy > 1, x > 0, y > 0): the sum = pi + tan^(-1)$\frac{(x+y}{(1-xy)}$), because the simple formula gives a value outside (-pi/2, pi/2) that must be adjusted. Case 3 (xy > 1, x < 0, y < 0): the sum = -pi + tan^(-1)$\frac{(x+y}{(1-xy)}$). The subtraction formula tan^(-1)(x) - tan^(-1)(y) = tan^(-1)$\frac{(x-y}{(1+xy)}$) when xy > -1. The condition on xy is the most tested trap in JEE. Classic example: tan^(-1)(2) + tan^(-1)(3) — since 23 = 6 > 1, the answer is pi + tan^(-1)$\frac{(2+3}{(1-6)}$) = pi + tan^(-1)(-1) = pi - pi/4 = 3pi/4. Students who ignore the condition get tan^(-1)(-1) = -pi/4, which is wrong.