Use the identity cos2θ+sin2θ≡1\cos^2\theta + \sin^2\theta \equiv 1cos2θ+sin2θ≡1 to prove that tan2θ≡sec2θ−1\tan^2\theta \equiv \sec^2\theta - 1tan2θ≡sec2θ−1.
Solve, for 0°≤θ≤360°0° \leq \theta \leq 360°0°≤θ≤360°, the equation
tan2θ+sec2θ+5secθ=2\tan^2\theta + \sec^2\theta + 5\sec\theta = 2tan2θ+sec2θ+5secθ=2
Give your answers to one decimal place.