In `DeltaABC,angleC=90^(@)andAB=c,BC=a, CA=b` , then the value of cosec B-cos A) is

Options

A. `(c^(2))/(ab)`
B. `(b^(2))/(ca)`
C. `(a^(2))/(bc)`
D. `(bc)/(a^(2))`

Answer

To solve the problem, we need to find the value of \( \csc B - \cos A \) in triangle \( \Delta ABC \) where \( \angle C = 90^\circ \), \( AB = c \), \( BC = a \), and \( CA = b \).

### Step-by-Step Solution:

1. **Identify the Triangle and Angles**:
- We have a right triangle \( \Delta ABC \) with \( \angle C = 90^\circ \).
- The sides are defined as follows: \( AB = c \) (hypotenuse), \( BC = a \) (opposite to angle A), and \( CA = b \) (opposite to angle B).

2. **Write the Definitions of \( \csc B \) and \( \cos A \)**:
- The cosecant of angle \( B \) is defined as:
\[ \csc B = \frac{1}{\sin B} = \frac{\text{hypotenuse}}{\text{opposite to } B} = \frac{c}{a} \]
- The cosine of angle \( A \) is defined as:
\[ \cos A = \frac{\text{adjacent to } A}{\text{hypotenuse}} = \frac{b}{c} \]

3. **Substitute the Values into the Expression**:
- We need to calculate \( \csc B - \cos A \):
\[ \csc B - \cos A = \frac{c}{a} - \frac{b}{c} \]

4. **Find a Common Denominator**:
- The common denominator for the two fractions is \( ac \):
\[ \csc B - \cos A = \frac{c^2}{ac} - \frac{b \cdot a}{ac} = \frac{c^2 - ab}{ac} \]

5. **Use the Pythagorean Theorem**:
- According to the Pythagorean theorem, we know that:
\[ c^2 = a^2 + b^2 \]
- Therefore, we can express \( c^2 - b^2 \) as:
\[ c^2 - b^2 = a^2 \]

6. **Substitute Back into the Expression**:
- Now substitute \( c^2 - b^2 \) into our expression:
\[ \csc B - \cos A = \frac{a^2}{ac} = \frac{a}{c} \]

### Final Answer:
The value of \( \csc B - \cos A \) is:
\[ \frac{a}{c} \]