If `y=(sin^(-1)(sinalphasinx)/(1-cosalphasinx))`, then `y'(0)` is equal to

Options

A. 1
B. `2tanalpha`
C. `((1)/(2))tanalpha`
D. `sinalpha`

Answer

To find \( y'(0) \) for the function

\[ y = \sin^{-1}\left(\frac{\sin \alpha \sin x}{1 - \cos \alpha \sin x}\right), \]

we will differentiate \( y \) with respect to \( x \) and then evaluate the derivative at \( x = 0 \).

### Step 1: Differentiate \( y \)

Using the chain rule and the quotient rule, we differentiate \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\sin^{-1}(u)\right) \quad \text{where} \quad u = \frac{\sin \alpha \sin x}{1 - \cos \alpha \sin x}. \]

From the chain rule, we know:

\[ \frac{d}{dx}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \]

### Step 2: Differentiate \( u \)

Now we need to differentiate \( u \) using the quotient rule:

\[ \frac{du}{dx} = \frac{(1 - \cos \alpha \sin x)(\sin \alpha \cos x) - (\sin \alpha \sin x)(-\cos \alpha \cos x)}{(1 - \cos \alpha \sin x)^2}. \]

### Step 3: Substitute \( u \) into the derivative

Now substituting \( u \) back into the derivative:

\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \]

### Step 4: Evaluate at \( x = 0 \)

Now we need to evaluate \( \frac{dy}{dx} \) at \( x = 0 \):

1. Calculate \( u \) at \( x = 0 \):

\[ u = \frac{\sin \alpha \sin(0)}{1 - \cos \alpha \sin(0)} = \frac{0}{1} = 0. \]

2. Calculate \( \sqrt{1 - u^2} \):

\[ \sqrt{1 - u^2} = \sqrt{1 - 0^2} = 1. \]

3. Calculate \( \frac{du}{dx} \) at \( x = 0 \):

Substituting \( x = 0 \) into \( \frac{du}{dx} \):

\[ \frac{du}{dx} = \frac{(1 - \cos \alpha \cdot 0)(\sin \alpha \cdot 1) - (\sin \alpha \cdot 0)(-\cos \alpha \cdot 1)}{(1 - \cos \alpha \cdot 0)^2} = \frac{1 \cdot \sin \alpha - 0}{1^2} = \sin \alpha. \]

### Step 5: Final calculation

Now substituting back into the derivative:

\[ \frac{dy}{dx}\bigg|_{x=0} = \frac{1}{1} \cdot \sin \alpha = \sin \alpha. \]

Thus,

\[ y'(0) = \sin \alpha. \]

### Final Answer

\[ y'(0) = \sin \alpha. \]

---