Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=a(theta-sintheta),y=b(1+costheta)`

Answer

To find \( \frac{dy}{dx} \) when \( x \) and \( y \) are connected parametrically by the equations \( x = a(\theta - \sin \theta) \) and \( y = b(1 + \cos \theta) \), we will differentiate both equations with respect to the parameter \( \theta \) and then use the chain rule to find \( \frac{dy}{dx} \).

### Step-by-Step Solution:

1. **Differentiate \( x \) with respect to \( \theta \)**:
\[ x = a(\theta - \sin \theta) \]
Differentiating both sides with respect to \( \theta \):
\[ \frac{dx}{d\theta} = a\left(1 - \cos \theta\right) \]

2. **Differentiate \( y \) with respect to \( \theta \)**:
\[ y = b(1 + \cos \theta) \]
Differentiating both sides with respect to \( \theta \):
\[ \frac{dy}{d\theta} = b\left(0 - \sin \theta\right) = -b \sin \theta \]

3. **Find \( \frac{dy}{dx} \)**:
Using the chain rule, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{-b \sin \theta}{a(1 - \cos \theta)} \]

4. **Simplifying the expression**:
We can simplify this expression further. Recall that \( 1 - \cos \theta = 2 \sin^2(\theta/2) \) and \( \sin \theta = 2 \sin(\theta/2) \cos(\theta/2) \). Thus:
\[ \frac{dy}{dx} = \frac{-b(2 \sin(\theta/2) \cos(\theta/2))}{a(2 \sin^2(\theta/2))} \]
Cancelling \( 2 \) from the numerator and denominator:
\[ \frac{dy}{dx} = \frac{-b \cos(\theta/2)}{a \sin(\theta/2)} \]
This can be rewritten using the cotangent function:
\[ \frac{dy}{dx} = -\frac{b}{a} \cot\left(\frac{\theta}{2}\right) \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{b}{a} \cot\left(\frac{\theta}{2}\right) \]