Show that: \[ \frac{1}{\sec \theta-1}+\frac{1}{\frac{\sec \theta+1}{\operatorname{cosec}^{2} \theta}}=2 \cos \theta+ \]

To show that \[ \frac{1}{\sec \theta - 1} + \frac{1}{\frac{\sec \theta + 1}{\operatorname{cosec}^{2} \theta}} = 2 \cos \theta, \] we will simplify the left-hand side step by step. ### Step 1: Rewrite the terms Recall that: \[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \operatorname{cosec}^2 \theta = \frac{1}{\sin^2 \theta}. \] Thus, we can rewrite the first term: \[ \frac{1}{\sec \theta - 1} = \frac{1}{\frac{1}{\cos \theta} - 1} = \frac{1}{\frac{1 - \cos \theta}{\cos \theta}} = \frac{\cos \theta}{1 - \cos \theta}. \] Now, for the second term: \[ \frac{1}{\frac{\sec \theta + 1}{\operatorname{cosec}^{2} \theta}} = \frac{\operatorname{cosec}^{2} \theta}{\sec \theta + 1} = \frac{\frac{1}{\sin^2 \theta}}{\frac{1}{\cos \theta} + 1} = \frac{1}{\sin^2 \theta} \cdot \frac{\cos \theta}{1 + \cos \theta} = \frac{\cos \theta}{\sin^2 \theta (1 + \cos \theta)}. \] ### Step 2: Combine the two fractions Now we have: \[ \frac{\cos \theta}{1 - \cos \theta} + \frac{\cos \theta}{\sin^2 \theta (1 + \cos \theta)}. \] To combine these fractions, we need a common denominator. The common denominator will be: \[ (1 - \cos \theta) \sin^2 \theta (1 + \cos \theta). \] Thus, we rewrite each term: 1. The first term becomes: \[ \frac{\cos \theta \cdot \sin^2 \theta (1 + \cos \theta)}{(1 - \cos \theta) \sin^2 \theta (1 + \cos \theta)} = \frac{\cos \theta \sin^2 \theta (1 + \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}. \] 2. The second term becomes: \[ \frac{\cos \theta (1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{\cos \theta (1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}. \] ### Step 3: Combine the numerators Now we can combine the numerators: \[ \frac{\cos \theta \sin^2 \theta (1 + \cos \theta) + \cos \theta (1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}. \] ### Step 4: Simplify the numerator The numerator simplifies to: \[ \cos \theta \left( \sin^2 \theta (1 + \cos \theta) + (1 - \cos \theta) \right). \] Expanding this gives: \[ \cos \theta \left( \sin^2 \theta + \sin^2 \theta \cos \theta + 1 - \cos \theta \right). \] Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can replace \(\sin^2 \theta\) with \(1 - \cos^2 \theta\): \[ \cos \theta \left( (1 - \cos^2 \theta) + \sin^2 \theta \cos \theta + 1 - \cos \theta \right). \] ### Step 5: Final simplification This simplifies to: \[ \cos \theta \left( 2 - \cos \theta \right). \] Thus, we have: \[ \frac{\cos \theta (2 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}. \] ### Step 6: Final result Now, we can simplify the left-hand side to show that it equals \(2 \cos \theta\): \[ \frac{\cos \theta (2 - \cos \theta)}{(1 - \cos^2 \theta)} = \frac{\cos \theta (2 - \cos \theta)}{\sin^2 \theta}. \] This leads us to: \[ \frac{2 \cos \theta}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} = 2 \cos \theta. \] Thus, we have shown that: \[ \frac{1}{\sec \theta - 1} + \frac{1}{\frac{\sec \theta + 1}{\operatorname{cosec}^{2} \theta}} = 2 \cos \theta. \] This completes the proof.