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 \(\sin^{-1}\)
We know that

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

Let \(a = \tan \theta\) and \(b = \tan \phi\). Then we can rewrite the left-hand side:

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

Thus, the equation becomes:

\[ 2\theta + 2\phi = 2\cot^{-1}\left(\frac{1}{x}\right). \]

### Step 2: Simplify the equation
Dividing both sides by 2 gives:

\[ \theta + \phi = \cot^{-1}\left(\frac{1}{x}\right). \]

### Step 3: Use the cotangent identity
Recall that

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

Thus, we have:

\[ \theta + \phi = \tan^{-1}(x). \]

### Step 4: Substitute back for \(\theta\) and \(\phi\)
Since \(\theta = \tan^{-1}(a)\) and \(\phi = \tan^{-1}(b)\), we can rewrite the equation as:

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

### Step 5: Use the addition formula for \(\tan^{-1}\)
Using the formula for the sum of arctangents:

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

we can equate:

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

### Step 6: Equate the arguments
Since the arctangent function is one-to-one, we can equate the arguments:

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

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

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