Problem 17 - Entrance Test

tan^2(θ) + cot^2(θ) = 2 => tan^2(θ) + 1/tan^2(θ) = 2 => tan^4(θ) - 2tan^2(θ) + 1 = 0 => (tan^2(θ) - 1)^2 = 0 => tan^2(θ) = 1 => tan(θ) = ±1. If tan(θ) = 1, then θ = nπ + π/4. If tan(θ) = -1, then θ = nπ - π/4. Therefore, the general solution is θ = nπ ± π/4.