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

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(y = a(\theta + \sin \theta)\) and \(x = a(1 + \cos \theta)\), we will differentiate both equations with respect to \(\theta\) and then use the chain rule.

### Step-by-Step Solution:

1. **Differentiate \(y\) with respect to \(\theta\)**:
\[ y = a(\theta + \sin \theta) \]
Differentiating \(y\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = a\left(\frac{d}{d\theta}(\theta) + \frac{d}{d\theta}(\sin \theta)\right) = a(1 + \cos \theta) \]

2. **Differentiate \(x\) with respect to \(\theta\)**:
\[ x = a(1 + \cos \theta) \]
Differentiating \(x\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = a\left(\frac{d}{d\theta}(1) + \frac{d}{d\theta}(\cos \theta)\right) = a(0 - \sin \theta) = -a \sin \theta \]

3. **Use the chain rule to find \(\frac{dy}{dx}\)**:
According to the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{a(1 + \cos \theta)}{-a \sin \theta} \]

4. **Simplify the expression**:
The \(a\) in the numerator and denominator cancels out:
\[ \frac{dy}{dx} = \frac{1 + \cos \theta}{-\sin \theta} \]
This can be rewritten as:
\[ \frac{dy}{dx} = -\frac{1 + \cos \theta}{\sin \theta} \]

5. **Further simplify using trigonometric identities**:
Recall that \(1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right)\) and \(\sin \theta = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)\):
\[ \frac{dy}{dx} = -\frac{2 \cos^2\left(\frac{\theta}{2}\right)}{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\cos\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} = -\cot\left(\frac{\theta}{2}\right) \]

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