If `x sin(a+y)+sina.cos(a+y)=0`, then prove that
`(dy)/(dx) = (sin^(2)(a+y))/(sina)`

Explanation

To solve the problem, we start with the given equation:

\[ x \sin(a + y) + \sin a \cos(a + y) = 0 \]

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate \( x \sin(a + y) \):

\[ x \sin(a + y) = -\sin a \cos(a + y) \]

### Step 2: Solving for \( x \)
Next, we can express \( x \) in terms of \( y \):

\[ x = -\frac{\sin a \cos(a + y)}{\sin(a + y)} \]

### Step 3: Differentiating Both Sides
Now, we differentiate both sides with respect to \( x \):

\[ \frac{d}{dx}(x) = \frac{d}{dx}\left(-\frac{\sin a \cos(a + y)}{\sin(a + y)}\right) \]

The left-hand side simplifies to:

\[ 1 \]

### Step 4: Applying the Quotient Rule on the Right Side
For the right-hand side, we will use the quotient rule. Let \( u = \sin a \cos(a + y) \) and \( v = \sin(a + y) \):

Using the quotient rule, we have:

\[ \frac{d}{dx}\left(-\frac{u}{v}\right) = -\frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step 5: Finding \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

1. **For \( u = \sin a \cos(a + y) \)**:
- Since \( \sin a \) is a constant, we have:
\[ \frac{du}{dx} = \sin a \cdot (-\sin(a + y)) \cdot \frac{dy}{dx} \]

2. **For \( v = \sin(a + y) \)**:
- We have:
\[ \frac{dv}{dx} = \cos(a + y) \cdot \frac{dy}{dx} \]

### Step 6: Substituting Back into the Quotient Rule
Now substituting \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) back into the quotient rule gives:

\[ 1 = -\frac{\sin(a + y)(-\sin a \sin(a + y) \frac{dy}{dx}) - \sin a \cos(a + y) \cos(a + y) \frac{dy}{dx}}{\sin^2(a + y)} \]

### Step 7: Simplifying the Equation
This simplifies to:

\[ 1 = \frac{\sin a \sin^2(a + y) \frac{dy}{dx} + \sin a \cos^2(a + y) \frac{dy}{dx}}{\sin^2(a + y)} \]

Factoring out \( \frac{dy}{dx} \):

\[ 1 = \frac{\sin a \frac{dy}{dx}}{\sin^2(a + y)} \]

### Step 8: Solving for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \]

### Conclusion
Thus, we have proven that:

\[ \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \]