Which of the following identieis, wherever defined, hold(s) good?

Options

A. `cot alpha-tan alpha=2cot 2alpha`
B. `tan(45^(@)+alpha)-tan(45^(@)-alpha)=2co sec 2alpha`
C. `tan(45^(@)+alpha)+tan(45^(@)-alpha)=2sec2alpha`
D. `tan alpha+cot alpha=2 tan 2alpha`.

Explanation

To solve the problem of determining which of the given identities hold true, we will analyze each option step by step.

### Option A: Prove that \( \cot \alpha - \tan \alpha = 2 \cot 2\alpha \)

1. **Start with the Left-Hand Side (LHS)**:
\[ \text{LHS} = \cot \alpha - \tan \alpha \]

2. **Rewrite in terms of sine and cosine**:
\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha}, \quad \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
Therefore,
\[ \text{LHS} = \frac{\cos \alpha}{\sin \alpha} - \frac{\sin \alpha}{\cos \alpha} \]

3. **Find a common denominator**:
\[ \text{LHS} = \frac{\cos^2 \alpha - \sin^2 \alpha}{\sin \alpha \cos \alpha} \]

4. **Use the identity for cosine of double angle**:
\[ \cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha \]
Thus,
\[ \text{LHS} = \frac{\cos 2\alpha}{\sin \alpha \cos \alpha} \]

5. **Use the identity for sine of double angle**:
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \]
Therefore,
\[ \text{LHS} = \frac{\cos 2\alpha}{\frac{1}{2} \sin 2\alpha} = \frac{2 \cos 2\alpha}{\sin 2\alpha} = 2 \cot 2\alpha \]

6. **Conclusion**:
\[ \text{LHS} = \text{RHS} \implies \cot \alpha - \tan \alpha = 2 \cot 2\alpha \quad \text{(True)} \]

### Option B: Prove that \( \tan(45^\circ + \alpha) - \tan(45^\circ - \alpha) = 2 \csc 2\alpha \)

1. **Start with the Left-Hand Side (LHS)**:
\[ \text{LHS} = \tan(45^\circ + \alpha) - \tan(45^\circ - \alpha) \]

2. **Use the tangent addition formula**:
\[ \tan(45^\circ + \alpha) = \frac{1 + \tan \alpha}{1 - \tan \alpha}, \quad \tan(45^\circ - \alpha) = \frac{1 - \tan \alpha}{1 + \tan \alpha} \]
Therefore,
\[ \text{LHS} = \frac{1 + \tan \alpha}{1 - \tan \alpha} - \frac{1 - \tan \alpha}{1 + \tan \alpha} \]

3. **Combine the fractions**:
\[ \text{LHS} = \frac{(1 + \tan \alpha)^2 - (1 - \tan \alpha)^2}{(1 - \tan \alpha)(1 + \tan \alpha)} \]

4. **Simplify the numerator**:
\[ (1 + \tan \alpha)^2 - (1 - \tan \alpha)^2 = 4 \tan \alpha \]
Thus,
\[ \text{LHS} = \frac{4 \tan \alpha}{1 - \tan^2 \alpha} \]

5. **Use the identity for tangent of double angle**:
\[ \tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} \]
Therefore,
\[ \text{LHS} = 2 \tan 2\alpha \]

6. **Conclusion**:
\[ \text{LHS} \neq 2 \csc 2\alpha \quad \text{(False)} \]

### Option C: Prove that \( \tan(45^\circ + \alpha) + \tan(45^\circ - \alpha) = 2 \csc 2\alpha \)

1. **Start with the Left-Hand Side (LHS)**:
\[ \text{LHS} = \tan(45^\circ + \alpha) + \tan(45^\circ - \alpha) \]

2. **Use the tangent addition formula**:
\[ \text{LHS} = \frac{1 + \tan \alpha}{1 - \tan \alpha} + \frac{1 - \tan \alpha}{1 + \tan \alpha} \]

3. **Combine the fractions**:
\[ \text{LHS} = \frac{(1 + \tan \alpha)^2 + (1 - \tan \alpha)^2}{(1 - \tan \alpha)(1 + \tan \alpha)} \]

4. **Simplify the numerator**:
\[ (1 + \tan \alpha)^2 + (1 - \tan \alpha)^2 = 2 + 2 \tan^2 \alpha \]
Thus,
\[ \text{LHS} = \frac{2(1 + \tan^2 \alpha)}{1 - \tan^2 \alpha} \]

5. **Use the identity for cosecant**:
\[ 1 + \tan^2 \alpha = \sec^2 \alpha \quad \text{and} \quad \sec^2 \alpha = \frac{1}{\cos^2 \alpha} \]
Therefore,
\[ \text{LHS} = \frac{2 \sec^2 \alpha}{1 - \tan^2 \alpha} \]

6. **Use the identity for sine of double angle**:
\[ \text{LHS} = 2 \csc 2\alpha \quad \text{(True)} \]

### Option D: Prove that \( \tan \alpha + \cot \alpha = 2 \tan 2\alpha \)

1. **Start with the Left-Hand Side (LHS)**:
\[ \text{LHS} = \tan \alpha + \cot \alpha \]

2. **Rewrite in terms of sine and cosine**:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha}, \quad \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \]
Therefore,
\[ \text{LHS} = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha} \]

3. **Use the Pythagorean identity**:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
Thus,
\[ \text{LHS} = \frac{1}{\sin \alpha \cos \alpha} \]

4. **Use the identity for sine of double angle**:
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \]
Therefore,
\[ \text{LHS} = \frac{2}{\sin 2\alpha} \]

5. **Conclusion**:
\[ \text{LHS} \neq 2 \tan 2\alpha \quad \text{(False)} \]

### Summary of Results:
- **Option A**: True
- **Option B**: False
- **Option C**: True
- **Option D**: False

### Final Answer:
The identities that hold true are **Option A** and **Option C**.