. If (1+tanθ)(1+tanϕ) = 2, then what is (θ + ϕ) equal to?

(1 + tanθ)(1 + tanϕ) = 2
1 (1 + tanϕ) + tanθ (1 + tanϕ) = 2
1 + tanϕ + tanθ + tanθ tanϕ = 2
tanϕ + tanθ = 1 - tanθ tanϕ
(tanθ + tanϕ) / (1 - tanθ tanϕ) = 1
tan(θ + ϕ) = 1
θ + ϕ = 1
Hence, θ + ϕ = nπ + (π/4), where n is an integer.

tan(a+b) = (tan(a) + tan(b)]/[1 - tan(a)tan(b)]

(1+tanA)(1+tanB) = 2
1+tanA +tanB+tanAtanB = 2
tanA + tanB + tanAtanB = 1
tanA + tanB = 1 - tanAtanB
[tanA + tanB]/[1 - tanAtanB] = 1 = tan(A+B)

tan(A+B) = 1
A+B = arctan(1) + kπ

Ans:
 (θ + ϕ) = π/4 + kπ, where k is any integer

On the interval [0, 2π):
 (θ + ϕ) = π/4, 5π/4