If ` xsin(a+y)+sina.cos(a+y)=0.` 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 \( x \) in terms of \( y \):

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

### Step 3: Simplifying the Expression
Using the trigonometric identity, we can simplify the expression:

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

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

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

### Step 5: Applying the Derivative of Cotangent
The derivative of \( \cot(u) \) is \( -\csc^2(u) \), so we apply the chain rule:

\[ \frac{d}{dy}(\cot(a + y)) = -\csc^2(a + y) \cdot \frac{d}{dy}(a + y) = -\csc^2(a + y) \cdot 1 = -\csc^2(a + y). \]

Thus, we have:

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

### Step 6: Simplifying Further
Since \( \csc^2(a + y) = \frac{1}{\sin^2(a + y)} \), we can write:

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

### Step 7: 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 proved that:

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