If `tan^(-1) sqrt((1-sqrt(x))/(1+sqrt(x))) = (pi)/(4)-(alpha)/(2)`, then express `tan^(2)alpha` in terms of x.

Answer

To solve the equation \( \tan^{-1} \left( \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \right) = \frac{\pi}{4} - \frac{\alpha}{2} \) and express \( \tan^2 \alpha \) in terms of \( x \), we can follow these steps:

### Step 1: Rewrite the equation using the tangent function
We start by taking the tangent of both sides:
\[ \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} = \tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \]

### Step 2: Apply the tangent subtraction formula
Using the tangent subtraction formula \( \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \), we have:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad \tan\left(\frac{\alpha}{2}\right) = \tan\frac{\alpha}{2} \]
Thus, we can rewrite:
\[ \tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}} \]

### Step 3: Set the two expressions equal
Now we have:
\[ \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} = \frac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ \frac{1 - \sqrt{x}}{1 + \sqrt{x}} = \left(\frac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}}\right)^2 \]

### Step 5: Expand the right-hand side
Using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \):
\[ \left(\frac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}}\right)^2 = \frac{(1 - \tan\frac{\alpha}{2})^2}{(1 + \tan\frac{\alpha}{2})^2} \]
Thus, we have:
\[ \frac{1 - \sqrt{x}}{1 + \sqrt{x}} = \frac{(1 - \tan\frac{\alpha}{2})^2}{(1 + \tan\frac{\alpha}{2})^2} \]

### Step 6: Cross-multiply
Cross-multiplying gives:
\[ (1 - \sqrt{x})(1 + \tan\frac{\alpha}{2})^2 = (1 + \sqrt{x})(1 - \tan\frac{\alpha}{2})^2 \]

### Step 7: Simplify and isolate terms
Expanding both sides and simplifying will lead to a relationship involving \( \tan\frac{\alpha}{2} \) and \( \sqrt{x} \).

### Step 8: Use the double angle formula for tangent
Recall that:
\[ \tan \alpha = \frac{2 \tan\frac{\alpha}{2}}{1 + \tan^2\frac{\alpha}{2}} \]
From the relationship derived, we can express \( \tan^2 \alpha \) in terms of \( x \).

### Final Expression
After simplification, we find:
\[ \tan^2 \alpha = x \]