If `int (1)/(sin (x -a ) cos (x -b)) dx = A log | (sin (x -a))/( cos (x -b ))| + B.` Then

Options

A. `A = cos (a -b) , B in R`
B. `A = sec (a -b) , B = C in R`
C. `A = sin (a-b), B in R`
D. `A =cosec (a-b) , B in R`

Answer

To solve the integral \( \int \frac{1}{\sin(x - a) \cos(x - b)} \, dx \) and find the value of \( A \) in the expression \( A \log \left| \frac{\sin(x - a)}{\cos(x - b)} \right| + B \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{\sin(x - a) \cos(x - b)} \, dx \]

### Step 2: Multiply and Divide by a Constant
To facilitate integration, we multiply and divide by \( \cos(a - b) \):
\[ I = \int \frac{\cos(a - b)}{\sin(x - a) \cos(x - b) \cos(a - b)} \, dx \]

### Step 3: Use Trigonometric Identity
Using the identity \( \cos(a - b) = \cos a \cos b + \sin a \sin b \), we can express the integral as:
\[ I = \int \frac{\cos(a - b)}{\sin(x - a) \cos(x - b)} \, dx \]

### Step 4: Split the Integral
We can split the integral into two parts:
\[ I = \frac{1}{\cos(a - b)} \left( \int \frac{\cos(x - b)}{\sin(x - a)} \, dx + \int \frac{\sin(x - a)}{\cos(x - b)} \, dx \right) \]

### Step 5: Evaluate Each Integral
Using the known integrals:
- \( \int \cot(x - a) \, dx = \log |\sin(x - a)| + C \)
- \( \int \tan(x - b) \, dx = -\log |\cos(x - b)| + C \)

We can write:
\[ I = \frac{1}{\cos(a - b)} \left( \log |\sin(x - a)| - \log |\cos(x - b)| \right) + C \]

### Step 6: Combine the Logarithms
Using properties of logarithms, we can combine the logs:
\[ I = \frac{1}{\cos(a - b)} \log \left| \frac{\sin(x - a)}{\cos(x - b)} \right| + C \]

### Step 7: Identify Constants
From the original expression \( A \log \left| \frac{\sin(x - a)}{\cos(x - b)} \right| + B \), we can identify:
\[ A = \frac{1}{\cos(a - b)} \]
and \( B \) is the constant of integration.

### Conclusion
Thus, the value of \( A \) is:
\[ A = \sec(a - b) \]