If A is not an integral multiple of `(pi)/(2)`, prove that
(i) `tan A + cot A= 2 cosec 2 A`
(ii) cot A - tan A = 2cot 2A`

Answer

To prove the given identities, we will start with the left-hand side (LHS) of each equation and simplify it to show that it equals the right-hand side (RHS).

### Part (i): Prove that \( \tan A + \cot A = 2 \csc 2A \)

1. **Start with LHS:**
\[ \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \]

2. **Combine the fractions:**
\[ = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} \]

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

4. **Express \( \sin A \cos A \) in terms of \( \sin 2A \):**
\[ \sin 2A = 2 \sin A \cos A \implies \sin A \cos A = \frac{1}{2} \sin 2A \]
Therefore,
\[ = \frac{1}{\frac{1}{2} \sin 2A} = \frac{2}{\sin 2A} \]

5. **Relate it to cosecant:**
\[ = 2 \csc 2A \]

6. **Conclusion:**
\[ \tan A + \cot A = 2 \csc 2A \]
Hence, the identity is proved.

### Part (ii): Prove that \( \cot A - \tan A = 2 \cot 2A \)

1. **Start with LHS:**
\[ \cot A - \tan A = \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} \]

2. **Combine the fractions:**
\[ = \frac{\cos^2 A - \sin^2 A}{\sin A \cos A} \]

3. **Use the double angle identity for cosine:**
\[ \cos^2 A - \sin^2 A = \cos 2A \]
Thus,
\[ = \frac{\cos 2A}{\sin A \cos A} \]

4. **Express \( \sin A \cos A \) in terms of \( \sin 2A \):**
\[ = \frac{\cos 2A}{\frac{1}{2} \sin 2A} = \frac{2 \cos 2A}{\sin 2A} \]

5. **Relate it to cotangent:**
\[ = 2 \cot 2A \]

6. **Conclusion:**
\[ \cot A - \tan A = 2 \cot 2A \]
Hence, the identity is proved.