If `sqrt(1-x^2) + sqrt(1-y^2)=a(x-y)`, 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-by-Step Solution:

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

2. **Substitute \(x = \sin \alpha\) and \(y = \sin \beta\):**
\[ \sqrt{1 - \sin^2 \alpha} + \sqrt{1 - \sin^2 \beta} = a(\sin \alpha - \sin \beta) \]
Using the identity \(1 - \sin^2 \theta = \cos^2 \theta\), we can rewrite the equation as:
\[ \cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta) \]

3. **Use trigonometric identities:**
Recall the identities:
\[ \sin C - \sin D = 2 \cos\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right) \]
\[ \cos C + \cos D = 2 \cos\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right) \]
Applying these identities:
\[ 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = a \cdot 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]

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

5. **Rearranging the equation:**
This can be rewritten as:
\[ \frac{\cos\left(\frac{\alpha - \beta}{2}\right)}{\sin\left(\frac{\alpha - \beta}{2}\right)} = a \]
Thus, we have:
\[ \cot\left(\frac{\alpha - \beta}{2}\right) = a \]

6. **Express \(\alpha - \beta\):**
From the cotangent, we can express:
\[ \frac{\alpha - \beta}{2} = \cot^{-1}(a) \implies \alpha - \beta = 2 \cot^{-1}(a) \]

7. **Differentiate both sides with respect to \(x\):**
Differentiating \(\alpha\) and \(\beta\):
\[ \frac{d}{dx}(\sin^{-1}(x)) - \frac{d}{dx}(\sin^{-1}(y)) = 0 \]
Using the chain rule:
\[ \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = 0 \]

8. **Solve for \(\frac{dy}{dx}\):**
Rearranging gives:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \]

9. **Final expression:**
Thus, we have shown that:
\[ \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \]