题目 Problem (a) Prove that (sinθ+cosθ)2−1cos2θ=2tanθ.\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 2\tan\theta.cos2θ(sinθ+cosθ)2−1=2tanθ. [3] (b) Hence solve the equation (sinθ+cosθ)2−1cos2θ=5tan3θ\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 5\tan^3\thetacos2θ(sinθ+cosθ)2−1=5tan3θ for −90∘<θ<90∘-90^\circ < \theta < 90^\circ−90∘<θ<90∘. [3] 题目中文翻译 (a) 证明 (sinθ+cosθ)2−1cos2θ=2tanθ\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 2\tan\thetacos2θ(sinθ+cosθ)2−1=2tanθ [3] (b) 由此解方程 (sinθ+cosθ)2−1cos2θ=5tan3θ\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 5\tan^3\thetacos2θ(sinθ+cosθ)2−1=5tan3θ 其中 −90∘<θ<90∘-90^\circ < \theta < 90^\circ−90∘<θ<90∘。 [3] 解答
