A triangle ABC is right angled at B. Find the value of `(sec A.sin C - tan A.tan C)/(sin B)`

Answer

To solve the problem, we need to evaluate the expression \((\sec A \cdot \sin C - \tan A \cdot \tan C) / \sin B\) for triangle ABC, which is right-angled at B.

### Step-by-Step Solution:

1. **Identify the Angles and Sides**:
In triangle ABC, since it is right-angled at B:
- \( \angle A + \angle C = 90^\circ \)
- Therefore, \( \sin C = \cos A \) and \( \tan C = \cot A \).

2. **Express Secant and Tangent**:
- The secant of angle A is given by:
\[ \sec A = \frac{1}{\cos A} \]
- The tangent of angle A is given by:
\[ \tan A = \frac{\sin A}{\cos A} \]
- The tangent of angle C can be expressed as:
\[ \tan C = \frac{\sin C}{\cos C} = \frac{\cos A}{\sin A} = \cot A \]

3. **Substituting Values into the Expression**:
Substitute \(\sec A\), \(\sin C\), \(\tan A\), and \(\tan C\) into the expression:
\[ \sec A \cdot \sin C = \frac{1}{\cos A} \cdot \cos A = 1 \]
\[ \tan A \cdot \tan C = \tan A \cdot \cot A = \frac{\sin A}{\cos A} \cdot \frac{\cos A}{\sin A} = 1 \]

4. **Putting It All Together**:
Now substitute these results back into the original expression:
\[ \frac{\sec A \cdot \sin C - \tan A \cdot \tan C}{\sin B} = \frac{1 - 1}{\sin B} = \frac{0}{\sin B} = 0 \]

### Final Answer:
The value of the expression \((\sec A \cdot \sin C - \tan A \cdot \tan C) / \sin B\) is \(0\).