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

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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.
$$ \frac{1}{\sec \theta - 1} + \frac{1}{\frac{\sec \theta + 1}{\operatorname{cosec}^{2} \theta}} = 2 \cos \theta $$

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