Consider a function `f(x)="sin"^(-1) (2x)/(1+x^(2))+"cos"^(-1) (1-x^(2))/(1+x^(2))+"tan"^(-1) (2x)/(1-x^(2))-atan^(-1)x(aepsilonR)`, the value of a if `f(x)=0` for all `x`:

Options

A. 6
B. `-6`
C. 2
D. `-2`

Explanation

To solve the problem, we need to find the value of \( a \) such that the function

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

is equal to 0 for all \( x \).

### Step 1: Substitute \( x = \tan(\theta) \)

Let \( x = \tan(\theta) \). Then we can rewrite the components of \( f(x) \):

- \( \frac{2x}{1+x^2} = \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \)
- \( \frac{1-x^2}{1+x^2} = \frac{1-\tan^2(\theta)}{1+\tan^2(\theta)} = \cos(2\theta) \)
- \( \frac{2x}{1-x^2} = \frac{2\tan(\theta)}{1-\tan^2(\theta)} = \tan(2\theta) \)

Thus, we can rewrite \( f(x) \) as:

\[ f(\tan(\theta)) = \sin^{-1}(\sin(2\theta)) + \cos^{-1}(\cos(2\theta)) + \tan^{-1}(\tan(2\theta)) - a \tan^{-1}(\tan(\theta)) \]

### Step 2: Simplify the function

Using the properties of inverse trigonometric functions, we have:

- \( \sin^{-1}(\sin(2\theta)) = 2\theta \)
- \( \cos^{-1}(\cos(2\theta)) = 2\theta \)
- \( \tan^{-1}(\tan(2\theta)) = 2\theta \)

So, we can simplify \( f(\tan(\theta)) \):

\[ f(\tan(\theta)) = 2\theta + 2\theta + 2\theta - a\theta = 6\theta - a\theta \]

### Step 3: Set the function equal to zero

Since we want \( f(x) = 0 \) for all \( x \), we set:

\[ 6\theta - a\theta = 0 \]

### Step 4: Factor out \( \theta \)

Factoring out \( \theta \) gives us:

\[ \theta(6 - a) = 0 \]

### Step 5: Solve for \( a \)

For this equation to hold for all \( \theta \), we must have:

\[ 6 - a = 0 \implies a = 6 \]

### Conclusion

Thus, the value of \( a \) is:

\[ \boxed{6} \]