Express: `cot^-1 (y/((1-x^2-y^2)))=2tan^-1 sqrt((3-4x^2)/(4x^2))- tan^-1 sqrt (3-4x^2)/x^2` as a rational integral equation in x and y.

Answer

To express the equation \( \cot^{-1} \left( \frac{y}{\sqrt{1 - x^2 - y^2}} \right) = 2 \tan^{-1} \left( \sqrt{\frac{3 - 4x^2}{4x^2}} \right) - \tan^{-1} \left( \frac{\sqrt{3 - 4x^2}}{x^2} \right) \) as a rational integral equation in \( x \) and \( y \), we can follow these steps:

### Step 1: Rewrite the Left-Hand Side (LHS)
We know that:
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]
Thus, we can rewrite the LHS:
\[ \cot^{-1} \left( \frac{y}{\sqrt{1 - x^2 - y^2}} \right) = \tan^{-1} \left( \frac{\sqrt{1 - x^2 - y^2}}{y} \right) \]

### Step 2: Rewrite the Right-Hand Side (RHS)
Using the double angle formula for tangent:
\[ 2 \tan^{-1}(x) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]
Let \( x = \sqrt{\frac{3 - 4x^2}{4x^2}} \). Then:
\[ 2 \tan^{-1} \left( \sqrt{\frac{3 - 4x^2}{4x^2}} \right) = \tan^{-1} \left( \frac{2\sqrt{\frac{3 - 4x^2}{4x^2}}}{1 - \frac{3 - 4x^2}{4x^2}} \right) \]
Simplifying the denominator:
\[ 1 - \frac{3 - 4x^2}{4x^2} = \frac{4x^2 - (3 - 4x^2)}{4x^2} = \frac{8x^2 - 3}{4x^2} \]
So, the RHS becomes:
\[ \tan^{-1} \left( \frac{2\sqrt{\frac{3 - 4x^2}{4x^2}} \cdot 4x^2}{8x^2 - 3} \right) - \tan^{-1} \left( \frac{\sqrt{3 - 4x^2}}{x^2} \right) \]

### Step 3: Combine the Two Terms on the RHS
Using the formula for the difference of two arctangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left( \frac{a - b}{1 + ab} \right) \]
Let \( a = \frac{2\sqrt{\frac{3 - 4x^2}{4x^2}} \cdot 4x^2}{8x^2 - 3} \) and \( b = \frac{\sqrt{3 - 4x^2}}{x^2} \). We can combine these into a single arctangent expression.

### Step 4: Set LHS Equal to RHS
Now we have:
\[ \tan^{-1} \left( \frac{\sqrt{1 - x^2 - y^2}}{y} \right) = \tan^{-1} \left( \text{combined expression from RHS} \right) \]

### Step 5: Eliminate the Arctangent
By taking the tangent of both sides, we get:
\[ \frac{\sqrt{1 - x^2 - y^2}}{y} = \text{combined expression from RHS} \]

### Step 6: Cross Multiply and Simplify
Cross-multiplying gives:
\[ \sqrt{1 - x^2 - y^2} = y \cdot \text{combined expression from RHS} \]
Squaring both sides leads to:
\[ 1 - x^2 - y^2 = y^2 \cdot \text{(RHS expression)}^2 \]

### Step 7: Rearranging
Rearranging gives us a rational integral equation in \( x \) and \( y \).

### Final Result
After simplification, we will arrive at a final equation of the form:
\[ 27y^2 = x^2(8 - x^2) \]