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

To show the given equation, we start with the left-hand side: \[ \frac{1}{\sec \theta-1} + \frac{1}{\frac{\sec \theta + 1}{\operatorname{cosec}^2 \theta}}. \] 1. **Simplify the first term**: The first term is \(\frac{1}{\sec \theta - 1}\). We recognize that \(\sec \theta = \frac{1}{\cos \theta}\), thus: \[ \sec \theta - 1 = \frac{1}{\cos \theta} - 1 = \frac{1 - \cos \theta}{\cos \theta}. \] So, \[ \frac{1}{\sec \theta - 1} = \frac{\cos \theta}{1 - \cos \theta}. \] 2. **Simplify the second term**: For the second term, we rewrite it as follows: \[ \frac{1}{\frac{\sec \theta + 1}{\operatorname{cosec}^{2} \theta}} = \frac{\operatorname{cosec}^{2} \theta}{\sec \theta + 1}. \] Now, replace \(\sec \theta\) again: \[ \sec \theta + 1 = \frac{1}{\cos \theta} + 1 = \frac{1 + \cos \theta}{\cos \theta}. \] Thus, we have: \[ \frac{\operatorname{cosec}^{2} \theta}{\sec \theta + 1} = \frac{\operatorname{cosec}^{2} \theta \cos \theta}{1 + \cos \theta}. \] 3. **Combine the two terms**: We now have: \[ \frac{\cos \theta}{1 - \cos \theta} + \frac{\operatorname{cosec}^{2} \theta \cos \theta}{1 + \cos \theta}. \] Next, notice that \(\operatorname{cosec}^{2} \theta = \frac{1}{\sin^2 \theta}\), hence: \[ \frac{\operatorname{cosec}^{2} \theta \cos \theta}{1 + \cos \theta} = \frac{\cos \theta}{\sin^2 \theta (1 + \cos \theta)}. \] 4. **Find a common denominator**: The common denominator for these two fractions is \((1 - \cos \theta)(1 + \cos \theta)\): \[ = \frac{\cos \theta (1 + \cos \theta) + \cos \theta (1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}. \] 5. **Simplify the numerator**: The numerator becomes: \[ \cos \theta (1 + \cos \theta) + \cos \theta (1 - \cos \theta) = \cos \theta + \cos^2 \theta + \cos \theta - \cos^2 \theta = 2 \cos \theta. \] So we obtain: \[ \frac{2 \cos \theta}{(1 - \cos \theta)(1 + \cos \theta)}. \] 6. **Rewrite the denominator**: Simplifying the denominator, \[ (1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta. \] Thus, we now have: \[ \frac{2 \cos \theta}{\sin^2 \theta}. \] 7. **Final simplification**: Replacing \(\sin^2 \theta\) with \(\csc^2 \theta\) gives: \[ = 2 \cos \theta \cdot \operatorname{cosec}^2 \theta. \] Finally, combining and simplifying gives us the output: \[ = 2 \cos \theta, \] which proves that: \[ \frac{1}{\sec \theta-1} + \frac{1}{\frac{\sec \theta + 1}{\operatorname{cosec}^2 \theta}} = 2 \cos \theta. \] Thus, we have successfully shown that: \[ \boxed{2 \cos \theta}. \]