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Let's break down and solve the problem in **Question number 3** step-by-step.

### Problem:
We are given:
\[
5 \cos^2(\theta) - 6 \sec^2(\theta) + 5 \sin^2(\theta) + 6 \tan^2(\theta)
\]
We need to simplify this expression and find its equivalent.

### Step-by-step simplification:

1. **Start with the given expression:**
   \[
   5 \cos^2(\theta) - 6 \sec^2(\theta) + 5 \sin^2(\theta) + 6 \tan^2(\theta)
   \]

2. **Use trigonometric identities:**
   - \(\sec^2(\theta) = 1 + \tan^2(\theta)\)
   - \(\sin^2(\theta) + \cos^2(\theta) = 1\)

3. **Substitute \(\sec^2(\theta)\) with \(1 + \tan^2(\theta)\):**
   \[
   5 \cos^2(\theta) - 6 (1 + \tan^2(\theta)) + 5 \sin^2(\theta) + 6 \tan^2(\theta)
   \]

4. **Expand the expression:**
   \[
   5 \cos^2(\theta) - 6 - 6 \tan^2(\theta) + 5 \sin^2(\theta) + 6 \tan^2(\theta)
   \]

5. **Combine like terms:**
   - The \(6 \tan^2(\theta)\) terms cancel out.
   \[
   5 \cos^2(\theta) + 5 \sin^2(\theta) - 6
   \]

6. **Use the Pythagorean identity** \(\sin^2(\theta) + \cos^2(\theta) = 1\):
   \[
   5(1) - 6 = 5 - 6 = -1
   \]

### Conclusion:
The simplified expression is:
\[
-1
\]

So the correct answer to **Question number 3** is:
\[
\boxed{-1}
\]

Would you like details or further explanations on any of these steps? 

Here are 5 related questions for further practice:
1. Simplify \(3 \sin^2(\theta) + 4 \cos^2(\theta)\).
2. What is the value of \(1 + \tan^2(\theta)\)?
3. How do you prove the identity \(\sec^2(\theta) - \tan^2(\theta) = 1\)?
4. Simplify \(\sin^4(\theta) + \cos^4(\theta)\).
5. Solve for \(\theta\) when \(\sec^2(\theta) = 4\).

**Tip**: Remember, Pythagorean identities like \(\sin^2(\theta) + \cos^2(\theta) = 1\) are fundamental in trigonometry and help simplify many expressions.

Which of the following is equivalent to 5 cos²(θ) − 6 sec²(θ) + 5 sin²(θ) + 6 tan²(θ)?

This problem simplifies a trigonometric expression using secant, sine, and tangent identities. It applies the Pythagorean identity sin²(θ) + cos²(θ) = 1 and sec²(θ) = 1 + tan²(θ) to derive the final result. Suitable for students in Grades 10-12.

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