`x=a(cos t + log tan (t/2)), y =a sin t` then find `dy/dx`

Answer

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

### Step 1: Differentiate \( x \) with respect to \( t \)

Given:
\[ x = a(\cos t + \log(\tan(t/2))) \]

We differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = a\left(-\sin t + \frac{d}{dt} \log(\tan(t/2))\right) \]

To differentiate \( \log(\tan(t/2)) \), we use the chain rule:
\[ \frac{d}{dt} \log(\tan(t/2)) = \frac{1}{\tan(t/2)} \cdot \frac{d}{dt}(\tan(t/2)) \]
Using the derivative of \( \tan(t/2) \):
\[ \frac{d}{dt}(\tan(t/2)) = \frac{1}{2} \sec^2(t/2) \]
Thus,
\[ \frac{d}{dt} \log(\tan(t/2)) = \frac{1}{\tan(t/2)} \cdot \frac{1}{2} \sec^2(t/2) = \frac{1}{2} \frac{\sec^2(t/2)}{\tan(t/2)} \]

Now substituting this back into the derivative of \( x \):
\[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{2} \frac{\sec^2(t/2)}{\tan(t/2)}\right) \]

### Step 2: Differentiate \( y \) with respect to \( t \)

Given:
\[ y = a \sin t \]

We differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = a \cos t \]

### Step 3: Find \( \frac{dy}{dx} \)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{a \cos t}{a\left(-\sin t + \frac{1}{2} \frac{\sec^2(t/2)}{\tan(t/2)}\right)} \]

The \( a \) cancels out:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{1}{2} \frac{\sec^2(t/2)}{\tan(t/2)}} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{1}{2} \frac{\sec^2(t/2)}{\tan(t/2)}} \]