If ` x =a cot theta , y=b cosec theta ,then (dy)/(dx)= `

Options

A. ` (-b)/( a ) cos theta `
B. ` (b)/(a) cos theta `
C. ` (-a)/(b) cos theta `
D. ` (a)/(b) cos theta `

Answer

To solve the problem where \( x = a \cot \theta \) and \( y = b \csc \theta \), we need to find \( \frac{dy}{dx} \). We'll use the chain rule to relate the derivatives with respect to \( \theta \).

### Step-by-Step Solution:

1. **Differentiate \( x \) with respect to \( \theta \)**:
\[ x = a \cot \theta \]
The derivative of \( \cot \theta \) is \( -\csc^2 \theta \). Therefore,
\[ \frac{dx}{d\theta} = a \cdot (-\csc^2 \theta) = -a \csc^2 \theta \]

2. **Differentiate \( y \) with respect to \( \theta \)**:
\[ y = b \csc \theta \]
The derivative of \( \csc \theta \) is \( -\csc \theta \cot \theta \). Therefore,
\[ \frac{dy}{d\theta} = b \cdot (-\csc \theta \cot \theta) = -b \csc \theta \cot \theta \]

3. **Use the chain rule to find \( \frac{dy}{dx} \)**:
By the chain rule,
\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} \]
We already have \( \frac{dy}{d\theta} \) and need to find \( \frac{d\theta}{dx} \). From \( \frac{dx}{d\theta} \):
\[ \frac{d\theta}{dx} = \frac{1}{\frac{dx}{d\theta}} = \frac{1}{-a \csc^2 \theta} = -\frac{1}{a \csc^2 \theta} \]

4. **Substituting into the chain rule**:
Now substitute \( \frac{dy}{d\theta} \) and \( \frac{d\theta}{dx} \):
\[ \frac{dy}{dx} = \left(-b \csc \theta \cot \theta\right) \cdot \left(-\frac{1}{a \csc^2 \theta}\right) \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{b \csc \theta \cot \theta}{a \csc^2 \theta} = \frac{b \cot \theta}{a \csc \theta} \]

5. **Using the identity \( \csc \theta = \frac{1}{\sin \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)**:
\[ \frac{dy}{dx} = \frac{b \frac{\cos \theta}{\sin \theta}}{a \frac{1}{\sin \theta}} = \frac{b \cos \theta}{a} \]

Thus, the final answer is:
\[ \frac{dy}{dx} = \frac{b \cos \theta}{a} \]