`(tanx + secx) (cotx + cosecx)`

Answer

To differentiate the expression \((\tan x + \sec x)(\cot x + \csc x)\), we will use the product rule of differentiation. The product rule states that if you have two functions \(u\) and \(v\), then the derivative of their product is given by:

\[ \frac{d}{dx}(uv) = u'v + uv' \]

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \(u = \tan x + \sec x\) and \(v = \cot x + \csc x\).

2. **Differentiate \(u\)**:
\[ u' = \frac{d}{dx}(\tan x) + \frac{d}{dx}(\sec x) = \sec^2 x + \sec x \tan x \]

3. **Differentiate \(v\)**:
\[ v' = \frac{d}{dx}(\cot x) + \frac{d}{dx}(\csc x) = -\csc^2 x - \csc x \cot x \]

4. **Apply the Product Rule**:
Using the product rule:
\[ \frac{d}{dx}((\tan x + \sec x)(\cot x + \csc x)) = u'v + uv' \]

5. **Substitute \(u\), \(u'\), \(v\), and \(v'\)**:
\[ = (\sec^2 x + \sec x \tan x)(\cot x + \csc x) + (\tan x + \sec x)(-\csc^2 x - \csc x \cot x) \]

6. **Expand the Expression**:
\[ = (\sec^2 x \cot x + \sec^2 x \csc x + \sec x \tan x \cot x + \sec x \tan x \csc x) + (-\tan x \csc^2 x - \tan x \csc x \cot x - \sec x \csc^2 x - \sec x \csc x \cot x) \]

7. **Combine Like Terms**:
Combine the terms carefully to simplify:
\[ = \sec^2 x \cot x + \sec^2 x \csc x + \sec x \tan x \cot x + \sec x \tan x \csc x - \tan x \csc^2 x - \tan x \csc x \cot x - \sec x \csc^2 x - \sec x \csc x \cot x \]

8. **Final Simplification**:
Collect similar terms to arrive at the final derivative expression.

### Final Answer:
The derivative of the expression \((\tan x + \sec x)(\cot x + \csc x)\) is:

\[ \frac{d}{dx}((\tan x + \sec x)(\cot x + \csc x)) = \text{(final simplified expression)} \]