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 \(x\):

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

### Step 2: Expressing \(x\)
Now, we can express \(x\) as:

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

### Step 3: Using Trigonometric Identities
Using the trigonometric identity \(\cot x = \frac{\cos x}{\sin x}\), we can rewrite \(x\):

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

### Step 4: Differentiating with Respect to \(y\)
Next, we differentiate both sides of the equation with respect to \(y\):

\[ \frac{dx}{dy} = -\sin a \cdot \frac{d}{dy}(\cot(a + y)) \]

Using the derivative of \(\cot x\), which is \(-\csc^2 x\):

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

### Step 5: Simplifying the Derivative
We can simplify \(\csc^2(a + y)\):

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

### Step 6: Finding \(\frac{dy}{dx}\)
To find \(\frac{dy}{dx}\), we take the reciprocal of \(\frac{dx}{dy}\):

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

### Conclusion
Thus, we have proven that:

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