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

Options

A. `-(pi)/(2)`
B. 0
C. `(pi)/(2)`
D. `pi`

Answer

To solve the expression \( 2\tan^{-1}\left(\frac{1-x^2}{2x}\right) + 2\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) for \( x \in (0, 1) \), we can follow these steps:

### Step 1: Rewrite the Expression
We can factor out the 2 from both terms:
\[ 2\left(\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) \]

### Step 2: Substitute \( x = \tan(\theta) \)
Let \( x = \tan(\theta) \). Then, we have:
\[ 1 - x^2 = 1 - \tan^2(\theta) = \frac{\cos(2\theta)}{\cos^2(\theta)} \]
and
\[ 1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta) \]

### Step 3: Substitute into the Inverse Functions
Now substituting into the inverse functions:
\[ \tan^{-1}\left(\frac{1-x^2}{2x}\right) = \tan^{-1}\left(\frac{\frac{\cos(2\theta)}{\cos^2(\theta)}}{2\tan(\theta)}\right) = \tan^{-1}\left(\frac{\cos(2\theta)}{2\sin(\theta)\cos(\theta)}\right) \]
This simplifies to:
\[ \tan^{-1}\left(\frac{\cos(2\theta)}{\sin(2\theta)}\right) = \tan^{-1}(\cot(2\theta)) = \frac{\pi}{2} - 2\theta \]

### Step 4: Simplify the Cosine Inverse
Now for the cosine inverse:
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \cos^{-1}\left(\frac{\frac{\cos(2\theta)}{\cos^2(\theta)}}{\sec^2(\theta)}\right) = \cos^{-1}(\cos(2\theta)) = 2\theta \]

### Step 5: Combine the Results
Now substituting back into our expression:
\[ 2\left(\frac{\pi}{2} - 2\theta + 2\theta\right) = 2\left(\frac{\pi}{2}\right) = \pi \]

### Final Result
Thus, the value of the original expression is:
\[ \pi \]