The solution the differential equation
`"cos x sin y dx" + "sin x cos y dy" =0` is

Options

A. `(sin x)/(sin y)=C`
B. `sin x sin y =C`
C. `sin x+sin y =C`
D. `cos x cos y =C`

Explanation

To solve the differential equation \( \cos x \sin y \, dx + \sin x \cos y \, dy = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ \cos x \sin y \, dx + \sin x \cos y \, dy = 0 \]
We can rearrange this to isolate \( dx \) and \( dy \):
\[ \cos x \sin y \, dx = -\sin x \cos y \, dy \]

### Step 2: Separating Variables
Next, we separate the variables by dividing both sides by \( \sin x \cos y \):
\[ \frac{\cos x}{\sin x} \, dx = -\frac{\cos y}{\sin y} \, dy \]
This simplifies to:
\[ \cot x \, dx = -\cot y \, dy \]

### Step 3: Integrating Both Sides
Now we integrate both sides:
\[ \int \cot x \, dx = -\int \cot y \, dy \]
Using the integral formula for \( \cot x \):
\[ \int \cot x \, dx = \log(\sin x) + C \]
Thus, we have:
\[ \log(\sin x) = -\log(\sin y) + C \]

### Step 4: Simplifying the Equation
We can rewrite the equation as:
\[ \log(\sin x) + \log(\sin y) = C \]
Using the property of logarithms that states \( \log a + \log b = \log(ab) \):
\[ \log(\sin x \sin y) = C \]

### Step 5: Exponentiating Both Sides
Exponentiating both sides to eliminate the logarithm gives:
\[ \sin x \sin y = e^C \]
Let \( e^C = k \), where \( k \) is a constant:
\[ \sin x \sin y = k \]

### Conclusion
The general solution of the differential equation is:
\[ \sin x \sin y = C \]
where \( C \) is a constant.