If `cos^(-1) ((1-a^(2))/(1+a^(2)))- cos^(-1) ((1-b^(2))/(1+b^(2))) = 2 tan^(-1) x`, then x is

Options

A. `(a-b)/(1+ab)`
B. `(b-a)/(1 + ab)`
C. `(a +b)/(1 - ab)`
D. None of these

Answer

To solve the equation

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

we will follow these steps:

### Step 1: Use the property of inverse cosine
We know that

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

Using this property, we can rewrite the left-hand side of the equation:

\[ \cos^{-1}\left(\frac{1-a^2}{1+a^2}\right) = 2 \tan^{-1}(a), \]
and
\[ \cos^{-1}\left(\frac{1-b^2}{1+b^2}\right) = 2 \tan^{-1}(b). \]

### Step 2: Substitute into the equation
Substituting these into our original equation gives:

\[ 2 \tan^{-1}(a) - 2 \tan^{-1}(b) = 2 \tan^{-1}(x). \]

### Step 3: Simplify the equation
Dividing the entire equation by 2, we have:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(x). \]

### Step 4: Use the property of inverse tangent
Using the property of inverse tangent, we know:

\[ \tan^{-1}(u) - \tan^{-1}(v) = \tan^{-1}\left(\frac{u-v}{1+uv}\right). \]

Applying this property to our equation, we can write:

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

### Step 5: Equate the arguments
Since both sides of the equation are equal, we can equate the arguments:

\[ x = \frac{a-b}{1+ab}. \]

### Conclusion
Thus, the value of \( x \) is

\[ \boxed{\frac{a-b}{1+ab}}. \]