If `x in (0, 1)`, then find the value of `tan^(-1) ((1 -x^(2))/(2x)) + cos^(-1) ((1 -x^(2))/(1 + x^(2)))`

Explanation

To solve the problem, we need to find the value of

\[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]

for \( x \in (0, 1) \).

### Step 1: Rewrite the cosine inverse term
We start by rewriting the cosine inverse term in a form that can be related to tangent. We know that

\[ \cos^{-1}(y) = \tan^{-1}\left(\frac{\sqrt{1 - y^2}}{y}\right) \]

for \( y = \frac{1 - x^2}{1 + x^2} \).

### Step 2: Calculate \( \sqrt{1 - y^2} \)
First, we need to calculate \( 1 - y^2 \):

\[ y^2 = \left(\frac{1 - x^2}{1 + x^2}\right)^2 = \frac{(1 - x^2)^2}{(1 + x^2)^2} \]

Now,

\[ 1 - y^2 = 1 - \frac{(1 - x^2)^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 - (1 - x^2)^2}{(1 + x^2)^2} \]

Calculating the numerator:

\[ (1 + x^2)^2 - (1 - x^2)^2 = (1 + 2x^2 + x^4) - (1 - 2x^2 + x^4) = 4x^2 \]

Thus,

\[ 1 - y^2 = \frac{4x^2}{(1 + x^2)^2} \]

### Step 3: Substitute back into the cosine inverse
Now substituting back into the expression for cosine inverse:

\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = \tan^{-1}\left(\frac{\sqrt{1 - y^2}}{y}\right) = \tan^{-1}\left(\frac{\sqrt{\frac{4x^2}{(1 + x^2)^2}}}{\frac{1 - x^2}{1 + x^2}}\right) \]

This simplifies to:

\[ \tan^{-1}\left(\frac{\frac{2x}{1 + x^2}}{\frac{1 - x^2}{1 + x^2}}\right) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]

### Step 4: Combine the terms
Now we can combine the two terms:

\[ \tan^{-1}\left(\frac{1 - x^2}{2x}\right) + \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]

Using the formula for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

where \( a = \frac{1 - x^2}{2x} \) and \( b = \frac{2x}{1 - x^2} \).

### Step 5: Calculate \( a + b \) and \( 1 - ab \)
Calculating \( a + b \):

\[ a + b = \frac{1 - x^2}{2x} + \frac{2x}{1 - x^2} = \frac{(1 - x^2)^2 + 4x^2}{2x(1 - x^2)} = \frac{1 - 2x^2 + x^4 + 4x^2}{2x(1 - x^2)} = \frac{1 + 2x^2 + x^4}{2x(1 - x^2)} \]

Calculating \( ab \):

\[ ab = \left(\frac{1 - x^2}{2x}\right) \left(\frac{2x}{1 - x^2}\right) = 1 \]

Thus, \( 1 - ab = 0 \).

### Step 6: Conclusion
Since \( 1 - ab = 0 \), the expression becomes undefined, which corresponds to \( \tan^{-1}(\infty) \), or \( \frac{\pi}{2} \).

Therefore, the final result is:

\[ \boxed{\frac{\pi}{2}} \]