if ` sin y = x sin ( alpha + y) and (dy)/(dx) = (a)/(x^2 + 2xb+1)` then

Options

A. `a=b`
B. ` a-b=1`
C. ` a+b=1`
D. ` a^2 +b^2 =1`

Answer

To solve the problem step by step, we start with the given equation and differentiate it with respect to \( x \).

### Step 1: Start with the given equation
The equation provided is:
\[ \sin y = x \sin(\alpha + y) \]

### Step 2: Apply the sine addition formula
Using the sine addition formula, we can expand the right-hand side:
\[ \sin(\alpha + y) = \sin \alpha \cos y + \cos \alpha \sin y \]
Substituting this back into the equation gives:
\[ \sin y = x (\sin \alpha \cos y + \cos \alpha \sin y) \]

### Step 3: Rearranging the equation
Rearranging the equation to isolate terms involving \( y \):
\[ \sin y - x \cos \alpha \sin y = x \sin \alpha \cos y \]
Factoring out \( \sin y \) on the left:
\[ \sin y (1 - x \cos \alpha) = x \sin \alpha \cos y \]

### Step 4: Divide by \( \cos y \)
To express \( \tan y \):
\[ \tan y = \frac{x \sin \alpha}{1 - x \cos \alpha} \]

### Step 5: Find \( y \)
Taking the inverse tangent:
\[ y = \tan^{-1}\left(\frac{x \sin \alpha}{1 - x \cos \alpha}\right) \]

### Step 6: Differentiate \( y \) with respect to \( x \)
Using the chain rule and the derivative of \( \tan^{-1}(u) \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{x \sin \alpha}{1 - x \cos \alpha}\right)^2} \cdot \frac{d}{dx}\left(\frac{x \sin \alpha}{1 - x \cos \alpha}\right) \]

### Step 7: Differentiate the fraction
Using the quotient rule:
Let \( u = x \sin \alpha \) and \( v = 1 - x \cos \alpha \):
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} = \sin \alpha \) and \( \frac{dv}{dx} = -\cos \alpha \):
\[ \frac{d}{dx}\left(\frac{x \sin \alpha}{1 - x \cos \alpha}\right) = \frac{(1 - x \cos \alpha)(\sin \alpha) - (x \sin \alpha)(-\cos \alpha)}{(1 - x \cos \alpha)^2} \]

### Step 8: Substitute back into the derivative
Substituting this back into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{(1 - x \cos \alpha)^2}{(1 + \left(\frac{x \sin \alpha}{1 - x \cos \alpha}\right)^2)} \cdot \left[\sin \alpha (1 - x \cos \alpha) + x \sin \alpha \cos \alpha\right] \]

### Step 9: Simplify the expression
After simplification, we find:
\[ \frac{dy}{dx} = \frac{a}{x^2 + 2xb + 1} \]
where \( a = \sin \alpha \) and \( b = -\cos \alpha \).

### Conclusion
Thus, we have:
\[ a = \sin \alpha, \quad b = -\cos \alpha \]