Prove that:(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.

Solution To prove: $$(\sec\theta - \cos\theta)(\cot\theta + \tan\theta) = \tan\theta \sec\theta$$ We know that $$\cot\theta = \frac{\cos\theta}{\sin\theta}$$ and $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$. Substituting these in the given equation, we get: $$(\sec\theta - \cos\theta)\left(\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\right) = \frac{\sin\theta}{\cos\theta} \cdot \frac{1}{\cos\theta}$$ Multiplying both sides by $$\frac{\sin\theta \cos\theta}{\cos\theta}$$, we get: $$(\sec\theta - \cos\theta)(\cos^2\theta + \sin^2\theta) = \sin\theta$$ Using the identity $$\cos^2\theta + \sin^2\theta = 1$$, we get: $$(\sec\theta - \cos\theta) = \frac{\sin\theta}{1} = \sin\theta$$ Multiplying both sides by $$\frac{\cos\theta}{\sin\theta}$$, we get: $$(\sec\theta - \cos\theta)(\cot\theta + \tan\theta) = \tan\theta \sec\theta$$ Therefore, we have shown that $$(\sec\theta - \cos\theta)(\cot\theta + \tan\theta) = \tan\theta \sec\theta$$.