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Solve all the questions one by one with the explaining all the tricks and concepts

Sure, let's go through each question one by one with detailed explanations.

### Question (i):
\[ \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} \]

We need to simplify the expression.

1. **Calculate \(\tan 30^\circ\)**:
   \[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]

2. **Substitute \(\tan 30^\circ\) into the expression**:
   \[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} \]

3. **Simplify the denominator**:
   \[ 1 + \left(\frac{1}{\sqrt{3}}\right)^2 = 1 + \frac{1}{3} = \frac{4}{3} \]

4. **Simplify the expression**:
   \[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2} \]

5. **Identify the result**:
   \[ \frac{\sqrt{3}}{2} = \cos 30^\circ \]

So, the correct option is:
\[ \boxed{\text{(B) cos } 60^\circ} \]

### Question (ii):
\[ \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} \]

1. **Calculate \(\tan 45^\circ\)**:
   \[ \tan 45^\circ = 1 \]

2. **Substitute \(\tan 45^\circ\) into the expression**:
   \[ \frac{1 - 1^2}{1 + 1^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \]

So, the correct option is:
\[ \boxed{\text{(D) 0}} \]

### Question (iii):
\[ \sin 2A = 2 \sin A \text{ is true when } A \text{ is:} \]

We know the double angle formula for sine:
\[ \sin 2A = 2 \sin A \cos A \]

For this to be true, \(\cos A\) must be 1, because:
\[ 2 \sin A \cos A = 2 \sin A \implies \cos A = 1 \]

This happens when:
\[ A = 0^\circ \]

So, the correct option is:
\[ \boxed{\text{(A) 0}^\circ} \]

### Question (iv):
\[ \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} \]

1. **Calculate \(\tan 30^\circ\)**:
   \[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]

2. **Substitute \(\tan 30^\circ\) into the expression**:
   \[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} \]

3. **Simplify the denominator**:
   \[ 1 - \left(\frac{1}{\sqrt{3}}\right)^2 = 1 - \frac{1}{3} = \frac{2}{3} \]

4. **Simplify the expression**:
   \[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{2} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

5. **Identify the result**:
   \[ \sqrt{3} = \tan 60^\circ \]

So, the correct option is:
\[ \boxed{\text{(C) tan } 60^\circ} \]

Would you like more details or have any questions?

Here are 5 questions you can ask:
1. How do you derive the double angle formula for sine?
2. What are some common trigonometric identities?
3. Can you explain the steps to solve another trigonometric expression?
4. What are the values of sine and cosine for standard angles?
5. How can we use trigonometric identities to simplify expressions?

**Tip:** Understanding fundamental trigonometric identities like \(\sin^2 \theta + \cos^2 \theta = 1\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) is crucial for solving many trigonometric problems.

Explore how to solve trigonometric expressions step-by-step with detailed explanations. This problem set covers fundamental trigonometric concepts, including identities and double angle formulas. Suitable for high school students.

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