Give a right angle triangle ABC with right angled at C. `angleC=90^(@)` and a,b,c are lengths of corresponding sides. `(b gt a)`.
`cos(A-B)` will be equal to

Options

A. cos2B
B. sin2A
C. cos2A
D. sinB

Answer

To solve the problem, we need to find the value of \( \cos(A - B) \) in a right triangle \( ABC \) where \( C \) is the right angle.

### Step-by-Step Solution:

1. **Understanding the Triangle**:
We have a right triangle \( ABC \) with \( \angle C = 90^\circ \). The angles \( A \) and \( B \) are the other two angles in the triangle.

2. **Using the Angle Sum Property**:
The sum of angles in a triangle is \( 180^\circ \). Therefore, we can write:
\[ A + B + C = 180^\circ \]
Since \( C = 90^\circ \), we have:
\[ A + B = 180^\circ - 90^\circ = 90^\circ \]
This implies:
\[ B = 90^\circ - A \]

3. **Substituting for \( B \)**:
Now, we can substitute \( B \) in the expression for \( \cos(A - B) \):
\[ \cos(A - B) = \cos(A - (90^\circ - A)) = \cos(A - 90^\circ + A) = \cos(2A - 90^\circ) \]

4. **Using the Cosine of a Difference**:
We can use the cosine subtraction formula:
\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]
Substituting \( B = 90^\circ - A \):
\[ \cos(A - B) = \cos A \cos(90^\circ - A) + \sin A \sin(90^\circ - A) \]

5. **Using Trigonometric Identities**:
We know that:
\[ \cos(90^\circ - A) = \sin A \quad \text{and} \quad \sin(90^\circ - A) = \cos A \]
Substituting these into the equation gives:
\[ \cos(A - B) = \cos A \cdot \sin A + \sin A \cdot \cos A \]
This simplifies to:
\[ \cos(A - B) = 2 \cos A \sin A \]

6. **Using the Double Angle Identity**:
We can use the double angle identity for sine:
\[ 2 \cos A \sin A = \sin(2A) \]
Thus, we have:
\[ \cos(A - B) = \sin(2A) \]

### Final Result:
The value of \( \cos(A - B) \) is:
\[ \cos(A - B) = \sin(2A) \]