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Let's solve all the questions from part 2.

### Question 2
Choose the correct option and explain your choice.

#### (i) \(\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}\)

First, find \(\tan 30^\circ\):

\[
\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\]

Now substitute:

\[
\frac{2 \cdot \frac{\sqrt{3}}{3}}{1 + \left(\frac{\sqrt{3}}{3}\right)^2} = \frac{\frac{2\sqrt{3}}{3}}{1 + \frac{1}{3}} = \frac{\frac{2\sqrt{3}}{3}}{\frac{4}{3}} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}
\]

Given options:

- (A) \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
- (B) \(\cos 60^\circ = \frac{1}{2}\)
- (C) \(\tan 60^\circ = \sqrt{3}\)
- (D) \(\sin 30^\circ = \frac{1}{2}\)

Thus, the correct answer is (A) \(\sin 60^\circ\).

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

First, find \(\tan 45^\circ\):

\[
\tan 45^\circ = 1 \quad \Rightarrow \quad \tan^2 45^\circ = 1
\]

Now substitute:

\[
\frac{1 - 1}{1 + 1} = \frac{0}{2} = 0
\]

Given options:

- (A) \(\tan 90^\circ\) (undefined)
- (B) 1
- (C) \(\sin 45^\circ = \frac{1}{\sqrt{2}}\)
- (D) 0

Thus, the correct answer is (D) 0.

#### (iii) \(\sin 2A = 2 \sin A \text{ is true when } A =\)

This statement can be verified using the double angle formula for sine:

\[
\sin 2A = 2 \sin A \cos A
\]

For this equation to be true:

\[
2 \sin A \cos A = 2 \sin A \quad \Rightarrow \quad \cos A = 1
\]

\(\cos A = 1\) when \(A = 0^\circ\).

Given options:

- (A) \(0^\circ\)
- (B) \(30^\circ\)
- (C) \(45^\circ\)
- (D) \(60^\circ\)

Thus, the correct answer is (A) \(0^\circ\).

#### (iv) \(\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ}\)

First, find \(\tan 30^\circ\):

\[
\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\]

Now substitute:

\[
\frac{2 \cdot \frac{\sqrt{3}}{3}}{1 - \left(\frac{\sqrt{3}}{3}\right)^2} = \frac{\frac{2\sqrt{3}}{3}}{1 - \frac{1}{3}} = \frac{\frac{2\sqrt{3}}{3}}{\frac{2}{3}} = \frac{2\sqrt{3}}{2} = \sqrt{3}
\]

Given options:

- (A) \(\cos 60^\circ = \frac{1}{2}\)
- (B) \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
- (C) \(\tan 60^\circ = \sqrt{3}\)
- (D) \(\sin 30^\circ = \frac{1}{2}\)

Thus, the correct answer is (C) \(\tan 60^\circ\).

Would you like to continue with more problems, or do you have any questions about these solutions so far?

Here are 5 questions you might want to ask:
1. How do you derive the double angle formulas in trigonometry?
2. What are some common mistakes to avoid when working with trigonometric identities?
3. Can you provide a step-by-step explanation for simplifying trigonometric expressions?
4. How do you solve trigonometric equations involving multiple angles?
5. What are the key differences between the basic and advanced trigonometric identities?

**Tip:** Always verify trigonometric identities by substituting known values and checking both sides of the equation for equality. This helps ensure your understanding and application of the identities are correct.

Explore solutions to trigonometric problems involving double angle formulas for sine, cosine, and tangent. This detailed guide covers multiple-choice questions and explanations for high school students.

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