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

Answer

To find the value of \(\frac{dy}{dx}\) given the parametric equations \(x = a(\cos t + \log \tan(t/2))\) and \(y = a \sin t\), we will follow these steps:

### Step 1: Differentiate \(y\) with respect to \(t\)
We start with the equation for \(y\):
\[ y = a \sin t \]
Differentiating \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = a \cos t \]

### Step 2: Differentiate \(x\) with respect to \(t\)
Next, we differentiate \(x\):
\[ x = a(\cos t + \log \tan(t/2)) \]
Using the product and chain rules, we differentiate:
\[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{\tan(t/2)} \cdot \frac{d}{dt}(\tan(t/2))\right) \]
Now, we need to differentiate \(\tan(t/2)\):
\[ \frac{d}{dt}(\tan(t/2)) = \sec^2(t/2) \cdot \frac{1}{2} = \frac{1}{2} \sec^2(t/2) \]
Thus, substituting this back into our equation for \(\frac{dx}{dt}\):
\[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{\tan(t/2)} \cdot \frac{1}{2} \sec^2(t/2)\right) \]
Using the identity \(\tan(t/2) = \frac{\sin(t/2)}{\cos(t/2)}\), we can simplify:
\[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{2} \cdot \frac{\sec^2(t/2)}{\tan(t/2)}\right) \]

### Step 3: Calculate \(\frac{dy}{dx}\)
Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \frac{a \cos t}{a\left(-\sin t + \frac{1}{2} \cdot \frac{\sec^2(t/2)}{\tan(t/2)}\right)} \]
The \(a\) cancels out:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{1}{2} \cdot \frac{\sec^2(t/2)}{\tan(t/2)}} \]

### Step 4: Simplify the expression
To simplify further, we can express \(\sec^2(t/2)\) and \(\tan(t/2)\) in terms of sine and cosine:
\[ \sec^2(t/2) = \frac{1}{\cos^2(t/2)}, \quad \tan(t/2) = \frac{\sin(t/2)}{\cos(t/2)} \]
Thus, we can rewrite:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{1}{2} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{\cos(t/2)}{\sin(t/2)}} \]
This can be further simplified based on the specific values of \(t\).

### Final Answer
The final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{1}{2} \cdot \frac{1}{\sin(t/2) \cos(t/2)}} \]