https://file.aiquickdraw.com/mathbot/file/1729856272801hh2992qt.jpg

Let's break down the mathematical content from the image.

### Problem 1 (First Triangle on the Left):
This is a right triangle with legs of lengths 5 and 12, and the hypotenuse is missing, but can be calculated using the Pythagorean theorem:

\[
\text{Hypotenuse} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]

Now we can calculate the trigonometric ratios for angle \(\theta\):
- \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13}\)
- \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{13}\)
- \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{5}\)
- \(\cot \theta = \frac{1}{\tan \theta} = \frac{5}{12}\)

### Problem 2 (Second Triangle on the Right):
This is another right triangle with legs of lengths 1 and 2, and the hypotenuse is missing but can be calculated:

\[
\text{Hypotenuse} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
\]

Now we calculate the trigonometric ratios for angle \(\theta\):
- \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\)
- \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\)
- \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{1} = 2\)
- \(\cot \theta = \frac{1}{\tan \theta} = \frac{1}{2}\)

### Problem 3 (Trigonometric Expression):
The next section asks to find:
- \(-\sin 30^\circ = -\frac{1}{2}\)

### Problem 4 (Trigonometric Identity Verification):
We need to verify this identity:
- \(\sin 30^\circ = \frac{1}{2}\), which is indeed true.

### Problem 5 (Trigonometric Expression for \(\tan(45^\circ)\)):
- \(\tan(45^\circ) = 1\), so the expression simplifies to:
  \[
  \sin(30^\circ) \times 1 = \frac{1}{2}
  \]

### Problem 6 (Simplifying an Expression):
Here we have:
\[
\sin 60^\circ \times \sin 30^\circ + \sin 60^\circ \times \cos 30^\circ
\]

- \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
- \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
- \(\sin 30^\circ = \frac{1}{2}\)

This simplifies to:
\[
\frac{\sqrt{3}}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} + \frac{3}{4} = \frac{\sqrt{3} + 3}{4}
\]

### Problem 7 (Expression Simplification):
This is more advanced, involving:
\[
\tan(20^\circ) \cdot \cot(20^\circ) + \sin(20^\circ) + \cos(20^\circ)
\]
We would need to calculate or simplify these trigonometric values.

### Problem 8 (Expression to Simplify):
\[
\frac{\tan(60^\circ) - \tan(20^\circ)}{1 + \tan(60^\circ) \cdot \tan(20^\circ)}
\]
Given:
- \(\tan(60^\circ) = \sqrt{3}\)

The expression involves specific values for \(\tan(20^\circ)\) and \(\tan(60^\circ)\).

Let me know if you need further steps or calculations for the unsimplified parts.

Would you like to delve into any of these further? Here are some related questions you could explore:

1. How do we derive the exact trigonometric ratios for special angles like 30°, 45°, and 60°?
2. How can the Pythagorean theorem be used in trigonometry?
3. What are the steps to verifying trigonometric identities?
4. How do you simplify trigonometric expressions involving multiple angles?
5. What are common trigonometric values for angles like 15° and 20°?

**Tip:** Always check if you can express trigonometric expressions in terms of known angles for easier simplification.

Solve the trigonometric ratios and expressions based on the given right triangles and simplify trigonometric identities.

Learn how to solve right triangles using the Pythagorean theorem and trigonometric ratios. Simplify trigonometric identities like tan(45°) and expressions involving sin(30°), cos(60°), and other common angles. Ideal for high school students in Grades 10-12.

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