If `u=cot^-1 sqrt(tanalpha)-tan^-1 sqrt(tan alpha),` then `tan(pi/4-u/2)` is equal to (a) `sqrt(tan alpha)` (b) `sqrt(cos alpha)` (c) `tan alpha` (d) `cot alpha`

Options

A. `sqrt(tan alpha)`
B. `sqrt(cot alpha)`
C. `tan alpha`
D. `cot alpha`

Explanation

To solve the problem, we need to evaluate \( \tan\left(\frac{\pi}{4} - \frac{u}{2}\right) \) where \( u = \cot^{-1}(\sqrt{\tan \alpha}) - \tan^{-1}(\sqrt{\tan \alpha}) \).

### Step-by-Step Solution:

1. **Define \( u \)**:
\[ u = \cot^{-1}(\sqrt{\tan \alpha}) - \tan^{-1}(\sqrt{\tan \alpha}) \]

2. **Let \( \sqrt{\tan \alpha} = \tan x \)**:
This substitution simplifies our expression for \( u \):
\[ u = \cot^{-1}(\tan x) - \tan^{-1}(\tan x) \]

3. **Use the identity for cotangent**:
Recall that \( \cot^{-1}(\tan x) = \frac{\pi}{2} - x \):
\[ u = \left(\frac{\pi}{2} - x\right) - x \]

4. **Simplify \( u \)**:
\[ u = \frac{\pi}{2} - 2x \]

5. **Rearranging for \( x \)**:
\[ 2x = \frac{\pi}{2} - u \implies x = \frac{\pi}{4} - \frac{u}{2} \]

6. **Finding \( \tan\left(\frac{\pi}{4} - \frac{u}{2}\right) \)**:
Using the tangent function:
\[ \tan\left(\frac{\pi}{4} - \frac{u}{2}\right) = \tan x \]

7. **Substituting back for \( \tan x \)**:
Since we defined \( \tan x = \sqrt{\tan \alpha} \):
\[ \tan\left(\frac{\pi}{4} - \frac{u}{2}\right) = \sqrt{\tan \alpha} \]

### Conclusion:
Thus, we find that:
\[ \tan\left(\frac{\pi}{4} - \frac{u}{2}\right) = \sqrt{\tan \alpha} \]

The correct answer is (a) \( \sqrt{\tan \alpha} \).