If `y=tan^(-1)((1-cos3x)/(sin3x))`, then `(dy)/(dx)=` . . .

Options

A. `-(3)/(2)`
B. `-(1)/(2)`
C. `(3)/(2)`
D. `(1)/(2)`

Explanation

To solve the problem, we need to find the derivative of the function \( y = \tan^{-1} \left( \frac{1 - \cos(3x)}{\sin(3x)} \right) \).

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We start with the expression for \( y \):
\[ y = \tan^{-1} \left( \frac{1 - \cos(3x)}{\sin(3x)} \right) \]

2. **Use Half-Angle Formula**:
We can use the half-angle identity for cosine:
\[ 1 - \cos(3x) = 2 \sin^2\left(\frac{3x}{2}\right) \]
Therefore, we can rewrite \( y \):
\[ y = \tan^{-1} \left( \frac{2 \sin^2\left(\frac{3x}{2}\right)}{\sin(3x)} \right) \]

3. **Use the Sine Double Angle Formula**:
The sine double angle formula states that:
\[ \sin(3x) = 2 \sin\left(\frac{3x}{2}\right) \cos\left(\frac{3x}{2}\right) \]
Thus, we can substitute this into our expression for \( y \):
\[ y = \tan^{-1} \left( \frac{2 \sin^2\left(\frac{3x}{2}\right)}{2 \sin\left(\frac{3x}{2}\right) \cos\left(\frac{3x}{2}\right)} \right) \]

4. **Simplify the Expression**:
The \( 2 \) cancels out:
\[ y = \tan^{-1} \left( \frac{\sin\left(\frac{3x}{2}\right)}{\cos\left(\frac{3x}{2}\right)} \right) \]
This simplifies to:
\[ y = \tan^{-1}(\tan\left(\frac{3x}{2}\right)) \]

5. **Final Simplification**:
Since \( \tan^{-1}(\tan(\theta)) = \theta \) for \( \theta \) in the principal range, we have:
\[ y = \frac{3x}{2} \]

6. **Differentiate with Respect to \( x \)**:
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{3}{2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{3}{2} \]