If `sin^(-1)((2a)/(1+a^(2)))+sin^(-1)((2b)/(1+b^(2)))= 2cot^(-1)((1)/(x))`, then x is equal to `[AA a, b in (0, 1)]`

Options

A. `(a-b)/(1+ab)`
B. `(b)/(1+ab)`
C. `(b)/(1-ab)`
D. `(a+b)/(1-ab)`

Answer

To solve the equation

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

we will follow these steps:

### Step 1: Use the identity for sine inverse
We know that

\[ \sin^{-1}(x) = 2 \tan^{-1}(x) \text{ for } x = \frac{2t}{1+t^2}. \]

Thus, we can rewrite the left side of the equation:

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

### Step 2: Substitute back into the equation
Substituting these into the equation gives:

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

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

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

### Step 4: Convert cotangent to tangent
Using the identity \(\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)\), we rewrite the right side:

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

### Step 5: Use the tangent addition formula
Using the formula for the sum of arctangents:

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

we equate:

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

### Step 6: Equate the arguments
Since the tangent functions are equal, we can set the arguments equal to each other:

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

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

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