`int[1+tanx.tan(x+alpha)]dx` is equal to

Options

A. (a) `cosalpha.ln|(sinx)/((sin(x+alpha)))|+C`
B. (b) `tanalpha.ln|(sinx)/(sin(x+alpha))|+C`
C. (c) `cotalpha.ln|(sec(x+alpha))/(secx)|+C`
D. (d) `cotalpha.ln|(cos(x+alpha))/(cosx)|+C`

Answer

To solve the integral \( I = \int (1 + \tan x \tan(x + \alpha)) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
Start by rewriting the integral:
\[ I = \int (1 + \tan x \tan(x + \alpha)) \, dx \]

### Step 2: Use the Identity
We can use the trigonometric identity:
\[ 1 + \tan a \tan b = \frac{\tan a - \tan b}{\tan a - b} \]
In our case, let \( a = x + \alpha \) and \( b = x \). Thus:
\[ 1 + \tan x \tan(x + \alpha) = \frac{\tan(x + \alpha) - \tan x}{\tan(x + \alpha) - x} \]

### Step 3: Substitute into the Integral
Substituting this back into the integral gives:
\[ I = \int \frac{\tan(x + \alpha) - \tan x}{\tan(x + \alpha) - x} \, dx \]

### Step 4: Simplify the Integral
This can be simplified further. We can separate the integral into two parts:
\[ I = \int \tan(x + \alpha) \, dx - \int \tan x \, dx \]

### Step 5: Integrate Each Part
We know that:
\[ \int \tan x \, dx = -\ln |\cos x| + C \]
Thus,
\[ \int \tan(x + \alpha) \, dx = -\ln |\cos(x + \alpha)| + C \]
So we can write:
\[ I = (-\ln |\cos(x + \alpha)|) - (-\ln |\cos x|) + C \]

### Step 6: Combine the Logarithms
Using the property of logarithms:
\[ \ln a - \ln b = \ln \left(\frac{a}{b}\right) \]
We have:
\[ I = \ln \left(\frac{|\cos x|}{|\cos(x + \alpha)|}\right) + C \]

### Step 7: Convert to Cotangent Form
Since \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), we can express the integral in terms of cotangent:
\[ I = \cot \alpha \ln \left(\frac{|\sec(x + \alpha)|}{|\sec x|}\right) + C \]

### Final Result
Thus, the integral evaluates to:
\[ I = \cot \alpha \ln \left(\frac{\sec(x + \alpha)}{\sec x}\right) + C \]

### Conclusion
This matches with option C from the given choices:
\[ \text{Option C: } \cot \alpha \ln \left(\frac{\sec(x + \alpha)}{\sec x}\right) + C \]