If `sin^(-1)((2a)/(1+a^(2)))-cos^(-1)((1-b^(2))/(1+b^2))=tan^(-1)((2x)/(1-x^(2)))` then what is the value of x?

Options

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

Answer

To solve the equation

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

we will use some properties and formulas of inverse trigonometric functions.

### Step 1: Use the formula for sine and cosine inverses

We know that:
\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}. \]
Thus, we can rewrite \(\cos^{-1}\) in terms of \(\sin^{-1}\):
\[ \cos^{-1}\left(\frac{1-b^2}{1+b^2}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{1-b^2}{1+b^2}\right). \]

### Step 2: Substitute into the equation

Substituting this into the original equation gives:
\[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) - \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1-b^2}{1+b^2}\right)\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right). \]
This simplifies to:
\[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{1-b^2}{1+b^2}\right) - \frac{\pi}{2} = \tan^{-1}\left(\frac{2x}{1-x^2}\right). \]

### Step 3: Use the formula for tangent inverse

Using the formula for the tangent of the sum of angles:
\[ \tan^{-1}(u) + \tan^{-1}(v) = \tan^{-1\left(\frac{u+v}{1-uv}\right)}, \]
we can express the left side in terms of tangent.

### Step 4: Apply the sine and cosine formulas

Using the known identities:
\[ \sin^{-1}(y) = 2 \tan^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right), \]
we can express the left-hand side in terms of tangent.

### Step 5: Set the equations equal

After simplifying, we find:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(x). \]

### Step 6: Use the tangent difference formula

Using the tangent difference formula:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right), \]
we can equate:
\[ \tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}(x). \]

### Step 7: Solve for \(x\)

This leads us to:
\[ x = \frac{a-b}{1+ab}. \]

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