The differential equation for which `sin^(-1) x + sin^(-1) y = c` is given by

Options

A. `sqrt(1-x^(2))dy+sqrt(1-y^(2))dx=0`
B. `sqrt(1-x^(2))dx+sqrt(1-y^(2))dy=0`
C. `sqrt(1-x^(2))dx-sqrt(1-y^(2))dy=0`
D. `sqrt(1-x^(2))dy-sqrt(1-y^(2))dx=0`

Answer

To find the differential equation for which \( \sin^{-1} x + \sin^{-1} y = c \), we can follow these steps:

### Step 1: Differentiate both sides
We start with the equation:
\[ \sin^{-1} x + \sin^{-1} y = c \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = \frac{d}{dx}(c) \]

### Step 2: Apply the differentiation formulas
Using the derivative of \( \sin^{-1} x \), which is \( \frac{1}{\sqrt{1 - x^2}} \):
\[ \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]

### Step 4: Isolate \( \frac{dy}{dx} \)
Now, we can isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \]

### Step 5: Rearranging to standard form
We can rearrange this into a more standard form:
\[ \sqrt{1 - y^2} \, dy + \sqrt{1 - x^2} \, dx = 0 \]

### Final Result
Thus, the differential equation corresponding to the given equation \( \sin^{-1} x + \sin^{-1} y = c \) is:
\[ \sqrt{1 - y^2} \, dy + \sqrt{1 - x^2} \, dx = 0 \]