If `sqrt(1-x^(2)) + sqrt(1-y^(2))=a(x-y)`, then prove that `(dy)/(dx) = sqrt((1-y^(2))/(1-x^(2)))`

Explanation

To prove that \(\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}\) given the equation \(\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)\), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \]

### Step 2: Substitute trigonometric identities
Let \(x = \sin \alpha\) and \(y = \sin \beta\). Then, we can rewrite the equation using the identities:
\[ \sqrt{1 - x^2} = \cos \alpha \quad \text{and} \quad \sqrt{1 - y^2} = \cos \beta \]
Thus, the equation becomes:
\[ \cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta) \]

### Step 3: Use sum-to-product identities
Using the sum-to-product identities for cosine:
\[ \cos \alpha + \cos \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
And for sine:
\[ \sin \alpha - \sin \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]
Substituting these into the equation gives:
\[ 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = a \cdot 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]

### Step 4: Cancel common terms
Assuming \(\cos\left(\frac{\alpha + \beta}{2}\right) \neq 0\), we can divide both sides by \(2 \cos\left(\frac{\alpha + \beta}{2}\right)\):
\[ \cos\left(\frac{\alpha - \beta}{2}\right) = a \sin\left(\frac{\alpha - \beta}{2}\right) \]

### Step 5: Express \(\alpha - \beta\)
This implies:
\[ \frac{\cos\left(\frac{\alpha - \beta}{2}\right)}{\sin\left(\frac{\alpha - \beta}{2}\right)} = a \]
Thus,
\[ \cot\left(\frac{\alpha - \beta}{2}\right) = a \]
From this, we can express \(\alpha - \beta\) as:
\[ \alpha - \beta = 2 \cot^{-1}(a) \]

### Step 6: Differentiate both sides
Now differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(\sin^{-1}(x) - \sin^{-1}(y)) = \frac{d}{dx}(2 \cot^{-1}(a)) \]
Using the derivatives:
\[ \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \]

### Step 7: Final result
Thus, we have:
\[ \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \]