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

Options

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

Explanation

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 double angle formula for tangent
We know that:
\[ 2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]
Thus, we can rewrite the equation as:
\[ \sin^{-1} \left( \frac{2a}{1 + a^2} \right) + \sin^{-1} \left( \frac{2b}{1 + b^2} \right) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]

### Step 2: Apply the sum of inverse sine formula
Using the identity for the sum of inverse sine functions:
\[ \sin^{-1} A + \sin^{-1} B = \sin^{-1} \left( A \sqrt{1 - B^2} + B \sqrt{1 - A^2} \right) \]
we can set \( A = \frac{2a}{1 + a^2} \) and \( B = \frac{2b}{1 + b^2} \).

### Step 3: Calculate \( A \sqrt{1 - B^2} + B \sqrt{1 - A^2} \)
First, we need to find \( \sqrt{1 - A^2} \) and \( \sqrt{1 - B^2} \):
\[ \sqrt{1 - A^2} = \sqrt{1 - \left( \frac{2a}{1 + a^2} \right)^2} = \sqrt{\frac{(1 + a^2)^2 - 4a^2}{(1 + a^2)^2}} = \sqrt{\frac{(1 - a^2)^2}{(1 + a^2)^2}} = \frac{1 - a^2}{1 + a^2} \]
Similarly,
\[ \sqrt{1 - B^2} = \sqrt{1 - \left( \frac{2b}{1 + b^2} \right)^2} = \frac{1 - b^2}{1 + b^2} \]

Now substituting back:
\[ \sin^{-1} \left( \frac{2a}{1 + a^2} \cdot \frac{1 - b^2}{1 + b^2} + \frac{2b}{1 + b^2} \cdot \frac{1 - a^2}{1 + a^2} \right) \]

### Step 4: Simplify the expression
This leads us to:
\[ \sin^{-1} \left( \frac{2a(1 - b^2) + 2b(1 - a^2)}{(1 + a^2)(1 + b^2)} \right) \]
This simplifies to:
\[ \sin^{-1} \left( \frac{2(a + b - ab)}{1 + a^2 + b^2 + ab} \right) \]

### Step 5: Set the two sides equal
Now we have:
\[ \sin^{-1} \left( \frac{2(a + b - ab)}{1 + a^2 + b^2 + ab} \right) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]

### Step 6: Equate the arguments of sine and tangent
From the identity:
\[ \sin^{-1}(y) = \tan^{-1}(z) \implies y = \frac{2z}{1 + z^2} \]
We can equate:
\[ \frac{2(a + b - ab)}{1 + a^2 + b^2 + ab} = \frac{2x}{1 + x^2} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 2(a + b - ab)(1 + x^2) = 2x(1 + a^2 + b^2 + ab) \]
Dividing both sides by 2:
\[ (a + b - ab)(1 + x^2) = x(1 + a^2 + b^2 + ab) \]
Expanding and rearranging terms leads to:
\[ x = \frac{a + b}{1 - ab} \]

### Final Answer
Thus, we find:
\[ x = \frac{a + b}{1 - ab} \]