If 3sinθ + 4cosθ = 5, then what is a value of 4tanθ + 3cotθ?

Given 3sinθ + 4cosθ = 5. Squaring: 9sin²θ + 24sinθcosθ + 16cos²θ = 25. Using sin²θ + cos²θ = 1: 9 + 7cos²θ + 24sinθcosθ = 25... Alternative approach: from 3sinθ + 4cosθ = 5 and the identity (3sinθ + 4cosθ)² + (4sinθ − 3cosθ)² = 25, we get 4sinθ − 3cosθ = 0, so tanθ = 3/4. Then 4tanθ + 3cotθ = 4(3/4) + 3(4/3) = 3 + 4 = 7.