Prove that: prove that 1/(secθ-tanθ)=secθ+tanθ

Solution We can start by using the definitions of secant and tangent: $$\frac{1}{\sec\theta - \tan\theta}$$ $$= \frac{1}{\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}}$$ $$= \frac{\cos\theta}{1 - \sin\theta}$$ Now, we can use the identity: $$1 - \sin^2\theta = \cos^2\theta$$ To rewrite the denominator: $$1 - \sin\theta = \cos^2\theta - \sin\theta\cos\theta$$ $$= \cos\theta(\cos\theta - \sin\theta)$$ So, we have: $$\frac{\cos\theta}{\cos\theta(\cos\theta - \sin\theta)}$$ $$= \frac{1}{\cos\theta - \sin\theta}$$ $$= \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \cdot \frac{\cos\theta + \sin\theta}{\cos\theta + \sin\theta}$$ $$= \frac{\cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta}{\cos^2\theta - \sin^2\theta}$$ $$= \frac{1 + 2\tan\theta\sec\theta + \tan^2\theta}{\sec^2\theta - \tan^2\theta}$$ $$= \frac{\sec\theta + \tan\theta)^2}{\sec^2\theta}$$ Therefore, we have shown that $$\frac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta$$.