Calculate the value of A, if :
(i) `(tan A - 1).("cosec 3A" - 1) = 0`
(ii) `cos 3A.(2 sin 2A - 1) = 0`

Answer

To solve the problem, we will break it down into two parts as given in the question.

### Part (i): Solve `(tan A - 1)(cosec 3A - 1) = 0`

1. **Set each factor to zero**:
\[ tan A - 1 = 0 \quad \text{or} \quad cosec 3A - 1 = 0 \]

2. **Solve the first equation**:
\[ tan A = 1 \]
The angle for which the tangent is 1 is:
\[ A = 45^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]

3. **Solve the second equation**:
\[ cosec 3A = 1 \]
This implies:
\[ sin 3A = 1 \]
The angle for which the sine is 1 is:
\[ 3A = 90^\circ + m \cdot 360^\circ \quad (m \in \mathbb{Z}) \]
Thus,
\[ A = 30^\circ + m \cdot 120^\circ \]

4. **Combine the results**:
From the first equation, we have \(A = 45^\circ\) and from the second equation, we have \(A = 30^\circ\).

### Part (ii): Solve `cos 3A(2 sin 2A - 1) = 0`

1. **Set each factor to zero**:
\[ cos 3A = 0 \quad \text{or} \quad 2 sin 2A - 1 = 0 \]

2. **Solve the first equation**:
\[ cos 3A = 0 \]
The angles for which the cosine is 0 are:
\[ 3A = 90^\circ + k \cdot 180^\circ \quad (k \in \mathbb{Z}) \]
Thus,
\[ A = 30^\circ + k \cdot 60^\circ \]

3. **Solve the second equation**:
\[ 2 sin 2A = 1 \quad \Rightarrow \quad sin 2A = \frac{1}{2} \]
The angles for which the sine is \( \frac{1}{2} \) are:
\[ 2A = 30^\circ + p \cdot 360^\circ \quad \text{or} \quad 2A = 150^\circ + p \cdot 360^\circ \quad (p \in \mathbb{Z}) \]
Thus,
\[ A = 15^\circ + p \cdot 180^\circ \quad \text{or} \quad A = 75^\circ + p \cdot 180^\circ \]

4. **Combine the results**:
From the first equation, we have \(A = 30^\circ\) and \(A = 90^\circ\) (not valid since it exceeds 90 degrees for our context). From the second equation, we have \(A = 15^\circ\) and \(A = 75^\circ\).

### Final Results:
- From Part (i): \(A = 45^\circ\) and \(A = 30^\circ\)
- From Part (ii): \(A = 30^\circ\) and \(A = 15^\circ\) and \(A = 75^\circ\)