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To evaluate the given limit:

\[
\lim_{\theta \to 0} \frac{1}{\theta} \left(\frac{1}{\sin \theta} - \frac{1}{\tan \theta}\right)
\]

First, simplify the expression inside the parentheses:

### Step 1: Simplify \(\frac{1}{\sin \theta} - \frac{1}{\tan \theta}\)
\[
\frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}
\]

Therefore:
\[
\frac{1}{\sin \theta} - \frac{1}{\tan \theta} = \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} = \frac{1 - \cos \theta}{\sin \theta}
\]

### Step 2: Substitute this back into the limit
\[
\lim_{\theta \to 0} \frac{1}{\theta} \cdot \frac{1 - \cos \theta}{\sin \theta}
\]

This can be rewritten as:
\[
\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}
\]

### Step 3: Use approximations for small \(\theta\)
As \(\theta \to 0\):
- \(\sin \theta \approx \theta\)
- \(1 - \cos \theta \approx \frac{\theta^2}{2}\)

Substitute these approximations:
\[
\lim_{\theta \to 0} \frac{\frac{\theta^2}{2}}{\theta \cdot \theta} = \lim_{\theta \to 0} \frac{\theta^2 / 2}{\theta^2} = \frac{1}{2}
\]

### Final Answer:
\[
\lim_{\theta \to 0} \frac{1}{\theta} \left(\frac{1}{\sin \theta} - \frac{1}{\tan \theta}\right) = \frac{1}{2}
\]

Would you like further clarification or details on any step?

Here are some related questions to consider:
1. How do we derive Taylor series for trigonometric functions?
2. What are other methods to evaluate limits involving trigonometric functions?
3. Why do small-angle approximations work effectively for limits?
4. How do we apply L'Hôpital's Rule in evaluating trigonometric limits?
5. What are some common pitfalls when simplifying trigonometric expressions?

**Tip:** When handling limits with trigonometric functions as \(\theta \to 0\), using series expansions or approximations can simplify the process and yield results quickly.

Evaluate the limit \( \lim_{\theta \to 0} \frac{1}{\theta} \left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right) \).

Learn how to evaluate the limit \( \lim_{\theta \to 0} \frac{1}{\theta} \left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right) \) using small-angle approximations and trigonometric identities. This solution provides a step-by-step approach, simplifying the expression and applying limit laws, making it suitable for high school students in Grades 11-12.

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