Use the identity cos2θ+sin2θ=1\cos^2\theta + \sin^2\theta = 1cos2θ+sin2θ=1 to prove that 1+tan2θ=sec2θ1 + \tan^2\theta = \sec^2\theta1+tan2θ=sec2θ
Solve, for 0≤θ≤3600 \leq \theta \leq 3600≤θ≤360, the equation,
2tan2θ−3secθ=0 2\tan^2\theta - 3\sec\theta = 0 2tan2θ−3secθ=0Give your answers to 1 decimal place.