A triangle ABC is right angled at B, find the value of `(secA.sinC-tanA-tanC)/(sinB)`

Answer

To solve the problem, we need to find the value of the expression \((\sec A \cdot \sin C - \tan A - \tan C) / \sin B\) for a right-angled triangle ABC, where the right angle is at B.

### Step-by-Step Solution:

1. **Identify the Angles and Sides:**
In triangle ABC, angle B is the right angle. Therefore, angles A and C are complementary, meaning \(A + C = 90^\circ\).

2. **Recall Trigonometric Identities:**
- \(\sec A = \frac{1}{\cos A} = \frac{\text{Hypotenuse}}{\text{Base}}\)
- \(\sin C = \frac{\text{Opposite to C}}{\text{Hypotenuse}}\)
- \(\tan A = \frac{\text{Opposite to A}}{\text{Base}}\)
- \(\tan C = \frac{\text{Opposite to C}}{\text{Base}}\)

3. **Assign Variables:**
Let:
- Hypotenuse = \(Z\)
- Base (adjacent to angle A) = \(X\)
- Opposite to angle A = \(Y\)

4. **Calculate Each Trigonometric Function:**
- \(\sec A = \frac{Z}{X}\)
- \(\sin C = \frac{Y}{Z}\)
- \(\tan A = \frac{Y}{X}\)
- \(\tan C = \frac{X}{Y}\)

5. **Substitute Values into the Expression:**
Substitute the values of \(\sec A\), \(\sin C\), \(\tan A\), and \(\tan C\) into the expression:
\[ \frac{\sec A \cdot \sin C - \tan A - \tan C}{\sin B} \]
becomes:
\[ \frac{\left(\frac{Z}{X} \cdot \frac{Y}{Z}\right) - \frac{Y}{X} - \frac{X}{Y}}{\sin B} \]

6. **Simplify the Expression:**
The expression simplifies to:
\[ \frac{\frac{Y}{X} - \frac{Y}{X} - \frac{X}{Y}}{\sin B} \]
Since \(\sin B = 1\) (because angle B is \(90^\circ\)):
\[ = \frac{-\frac{X}{Y}}{1} = -\frac{X}{Y} \]

7. **Final Result:**
Thus, the value of the expression is:
\[ -\frac{X}{Y} \]