If sinθ + cosθ = 2a and tanθ + cotθ = b, then find a

Solution

Start with sinθ + cosθ = 2a. Squaring both sides gives: sin²θ + cos²θ + 2sinθcosθ = 4a². Using sin²θ + cos²θ = 1: 1 + 2sinθcosθ = 4a². Then, we know from tanθ + cotθ = b, that: tanθ + cotθ = sinθ/cosθ + cosθ/sinθ = (sin²θ + cos²θ)/sinθcosθ = 1/sinθcosθ = b. Therefore: sinθcosθ = 1/b. Substituting back, we have: 1 + 2(1/b) = 4a². Rearranging gives: 4a² = 1 + 2/b. Thus, we can solve for a in terms of b: a = √(1 + 2/b)/2. Correct answer : b) √(1 + 2/b)/2.