To simplify the given expression:

$\frac{\csc \theta (\sec^2 \theta - 1)}{\sec \theta}$

Step-by-step Solution:

1. Recall a Pythagorean Identity:

$\sec^2 \theta - 1 = \tan^2 \theta$

This identity comes directly from $\sec^2 \theta = 1 + \tan^2 \theta$.

2. Substitute into the Expression:

$\frac{\csc \theta (\sec^2 \theta - 1)}{\sec \theta} \implies \frac{\csc \theta (\tan^2 \theta)}{\sec \theta}$

3. Rewrite Trigonometric Functions:

$\csc \theta = \frac{1}{\sin \theta}, \quad \sec \theta = \frac{1}{\cos \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}$

Thus:

$\tan^2 \theta = \left( \frac{\sin \theta}{\cos \theta} \right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta}$

4. Substitute These Values:

$\frac{\csc \theta (\tan^2 \theta)}{\sec \theta} = \frac{\left( \frac{1}{\sin \theta} \right) \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)}{\frac{1}{\cos \theta}}$

5. Simplify the Expression:

• Combine terms in the numerator:

$\frac{1}{\sin \theta} \cdot \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta}{\sin \theta \cos^2 \theta} = \frac{\sin \theta}{\cos^2 \theta}$

• Divide by $\frac{1}{\cos \theta}$ in the denominator:

$\frac{\frac{\sin \theta}{\cos^2 \theta}}{\frac{1}{\cos \theta}} = \frac{\sin \theta}{\cos^2 \theta} \cdot \cos \theta = \frac{\sin \theta \cos \theta}{\cos^2 \theta}$

Simplify further:

$\frac{\sin \theta \cos \theta}{\cos^2 \theta} = \frac{\sin \theta}{\cos \theta}$

6. Recognize the Simplified Result:

$\frac{\sin \theta}{\cos \theta} = \tan \theta$

Final Answer:

The expression is equivalent to (C) $\tan \theta$.

Do you need further clarification or additional steps? Here are 5 related questions:

1. How do you derive the identity $\sec^2 \theta - 1 = \tan^2 \theta$?
2. What are the basic Pythagorean identities in trigonometry?
3. How do you simplify expressions involving $\csc \theta$ and $\sec \theta$?
4. Can you explain how to rewrite trigonometric functions in terms of sine and cosine?
5. How do you simplify complex trigonometric fractions?

Tip: Always simplify trigonometric expressions using fundamental identities like Pythagorean identities and definitions of functions!