Solve ` cot theta = sin 2 theta ` by substituting ` sin 2 theta =( 2 tan theta )/( 1+tan^(2) theta )` and again by substituting ` sin 2 theta = 2 sin theta * cos theta ` and check whether the two answer are same or not .

Answer

To solve the equation \( \cot \theta = \sin 2\theta \) using two different substitutions, we will follow these steps:

### Step 1: Using the substitution \( \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \)

1. Start with the equation:
\[ \cot \theta = \sin 2\theta \]
Substitute \( \sin 2\theta \):
\[ \cot \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \]

2. Rewrite \( \cot \theta \) in terms of \( \tan \theta \):
\[ \frac{1}{\tan \theta} = \frac{2 \tan \theta}{1 + \tan^2 \theta} \]

3. Cross-multiply to eliminate the fractions:
\[ 1 + \tan^2 \theta = 2 \tan^2 \theta \]

4. Rearranging gives:
\[ 1 = 2 \tan^2 \theta - \tan^2 \theta \]
\[ 1 = \tan^2 \theta \]

5. Take the square root of both sides:
\[ \tan \theta = \pm 1 \]

6. The general solutions for \( \tan \theta = 1 \) and \( \tan \theta = -1 \) are:
\[ \theta = n\pi + \frac{\pi}{4} \quad \text{and} \quad \theta = n\pi - \frac{\pi}{4} \]
where \( n \) is any integer.

### Step 2: Using the substitution \( \sin 2\theta = 2 \sin \theta \cos \theta \)

1. Start with the equation:
\[ \cot \theta = \sin 2\theta \]
Substitute \( \sin 2\theta \):
\[ \cot \theta = 2 \sin \theta \cos \theta \]

2. Rewrite \( \cot \theta \) in terms of \( \sin \theta \) and \( \cos \theta \):
\[ \frac{\cos \theta}{\sin \theta} = 2 \sin \theta \cos \theta \]

3. Cross-multiply to eliminate the fractions:
\[ \cos \theta = 2 \sin^2 \theta \cos \theta \]

4. If \( \cos \theta \neq 0 \), we can divide both sides by \( \cos \theta \):
\[ 1 = 2 \sin^2 \theta \]

5. Rearranging gives:
\[ \sin^2 \theta = \frac{1}{2} \]

6. Take the square root of both sides:
\[ \sin \theta = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

7. The general solutions for \( \sin \theta = \frac{\sqrt{2}}{2} \) and \( \sin \theta = -\frac{\sqrt{2}}{2} \) are:
\[ \theta = n\pi + \frac{\pi}{4} \quad \text{and} \quad \theta = n\pi - \frac{\pi}{4} \]
where \( n \) is any integer.

### Conclusion

Both methods yield the same general solutions:
- From the first substitution: \( \theta = n\pi + \frac{\pi}{4} \) and \( \theta = n\pi - \frac{\pi}{4} \)
- From the second substitution: \( \theta = n\pi + \frac{\pi}{4} \) and \( \theta = n\pi - \frac{\pi}{4} \)

Thus, the solutions are consistent.