The value of `int {1+tan x*tan (x + A)}\ dx` is

Options

A. `cot A * log |(sec x)/(sec(x+A))|+C`
B. `tan A * log |sec(x+A)|+C`
C. `cotA*log|(sec(x+A))/(sec(x))|+C`
D. None of these

Answer

To solve the integral \( \int (1 + \tan x \tan(x + A)) \, dx \), we will follow these steps:

### Step 1: Use the trigonometric identity
We know that:
\[ \tan(x + A) = \frac{\tan x + \tan A}{1 - \tan x \tan A} \]
Thus, we can rewrite \( \tan x \tan(x + A) \):
\[ \tan x \tan(x + A) = \tan x \cdot \frac{\tan x + \tan A}{1 - \tan x \tan A} \]

### Step 2: Simplify the expression
Now, we can express the integrand:
\[ 1 + \tan x \tan(x + A) = 1 + \frac{\tan^2 x + \tan x \tan A}{1 - \tan x \tan A} \]
This simplifies to:
\[ \frac{(1 - \tan x \tan A) + (\tan^2 x + \tan x \tan A)}{1 - \tan x \tan A} = \frac{1 + \tan^2 x}{1 - \tan x \tan A} \]

### Step 3: Use the identity \( 1 + \tan^2 x = \sec^2 x \)
Thus, we have:
\[ 1 + \tan x \tan(x + A) = \frac{\sec^2 x}{1 - \tan x \tan A} \]

### Step 4: Rewrite the integral
Now we can rewrite the integral:
\[ \int (1 + \tan x \tan(x + A)) \, dx = \int \frac{\sec^2 x}{1 - \tan x \tan A} \, dx \]

### Step 5: Use substitution
Let \( u = \tan x \), then \( du = \sec^2 x \, dx \). The integral becomes:
\[ \int \frac{1}{1 - u \tan A} \, du \]

### Step 6: Integrate using the formula for arctangent
The integral \( \int \frac{1}{1 - au} \, du = -\frac{1}{a} \ln |1 - au| + C \) gives us:
\[ -\frac{1}{\tan A} \ln |1 - \tan x \tan A| + C \]

### Step 7: Final expression
Thus, the final value of the integral is:
\[ -\frac{1}{\tan A} \ln |1 - \tan x \tan A| + C \]