Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=asintheta,y=a(costheta+log"tan"theta/2)`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = a \sin \theta\) and \(y = a \left(\cos \theta + \log \tan \frac{\theta}{2}\right)\), we will use the chain rule for parametric differentiation.

### Step-by-Step Solution:

1. **Differentiate \(y\) with respect to \(\theta\)**:
\[ y = a \left(\cos \theta + \log \tan \frac{\theta}{2}\right) \]
We will differentiate \(y\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = a \left(-\sin \theta + \frac{d}{d\theta} \log \tan \frac{\theta}{2}\right) \]
Using the chain rule for the logarithm:
\[ \frac{d}{d\theta} \log u = \frac{1}{u} \cdot \frac{du}{d\theta} \]
where \(u = \tan \frac{\theta}{2}\). Therefore,
\[ \frac{du}{d\theta} = \frac{1}{2} \sec^2 \frac{\theta}{2} \]
Thus,
\[ \frac{d}{d\theta} \log \tan \frac{\theta}{2} = \frac{1}{\tan \frac{\theta}{2}} \cdot \frac{1}{2} \sec^2 \frac{\theta}{2} \]

2. **Combine the derivatives**:
\[ \frac{dy}{d\theta} = a \left(-\sin \theta + \frac{1}{\tan \frac{\theta}{2}} \cdot \frac{1}{2} \sec^2 \frac{\theta}{2}\right) \]

3. **Differentiate \(x\) with respect to \(\theta\)**:
\[ x = a \sin \theta \]
Differentiating \(x\) gives:
\[ \frac{dx}{d\theta} = a \cos \theta \]

4. **Find \(\frac{dy}{dx}\)**:
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \left(-\sin \theta + \frac{1}{\tan \frac{\theta}{2}} \cdot \frac{1}{2} \sec^2 \frac{\theta}{2}\right)}{a \cos \theta} \]
The \(a\) cancels out:
\[ \frac{dy}{dx} = \frac{-\sin \theta + \frac{1}{\tan \frac{\theta}{2}} \cdot \frac{1}{2} \sec^2 \frac{\theta}{2}}{\cos \theta} \]

5. **Simplify the expression**:
Recognizing that \(\tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\) and \(\sec^2 \frac{\theta}{2} = \frac{1}{\cos^2 \frac{\theta}{2}}\), we can further simplify the expression if necessary.

### Final Result:
Thus, the derivative \(\frac{dy}{dx}\) is given by:
\[ \frac{dy}{dx} = \frac{-\sin \theta + \frac{1}{2} \cdot \frac{\sec^2 \frac{\theta}{2}}{\tan \frac{\theta}{2}}}{\cos \theta} \]