`sin^(-1)((2a)/(1+a^(2)))+sin^(-1)((2b)/(1+b^(2)))=2tan^(-1)x`, then x is equal to

Options

A. A) `(a-b)/(1+ab)`
B. B) `(a+b)/(1-ab)`
C. C) `(2ab)/(a+b)`
D. D) `(a-b)/(1+ab)`

Answer

To solve the equation \( \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}(x) \), we can follow these steps:

### Step 1: Use the property of inverse sine
We know that:
\[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) = 2\tan^{-1}(a) \]
and
\[ \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}(b) \]
Thus, we can rewrite the left-hand side of the equation:
\[ 2\tan^{-1}(a) + 2\tan^{-1}(b) = 2\tan^{-1}(x) \]

### Step 2: Simplify the equation
Dividing both sides by 2 gives:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}(x) \]

### Step 3: Use the formula for the sum of inverse tangents
We can use the formula:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
This leads us to:
\[ \tan^{-1}\left(\frac{a+b}{1-ab}\right) = \tan^{-1}(x) \]

### Step 4: Equate the arguments
Since the inverse tangent function is one-to-one, we can equate the arguments:
\[ x = \frac{a+b}{1-ab} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{a+b}{1-ab}} \]