Statement Prove the identity secθ−cosθtanθ≡sinθ.\frac{\sec\theta-\cos\theta}{\tan\theta}\equiv \sin\theta.tanθsecθ−cosθ≡sinθ. Solution Start from the numerator: secθ−cosθ=1cosθ−cosθ=1−cos2θcosθ=sin2θcosθ.\sec\theta-\cos\theta=\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta}.secθ−cosθ=cosθ1−cosθ=cosθ1−cos2θ=cosθsin2θ. Divide by tanθ=sinθcosθ\tan\theta=\dfrac{\sin\theta}{\cos\theta}tanθ=cosθsinθ: sin2θcosθ⋅cosθsinθ=sinθ.\frac{\sin^2\theta}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}=\sin\theta.cosθsin2θ⋅sinθcosθ=sinθ. Hence secθ−cosθtanθ=sinθ\dfrac{\sec\theta-\cos\theta}{\tan\theta}=\boxed{\sin\theta}tanθsecθ−cosθ=sinθ. ■\blacksquare■