If ` x=a [ cos t +log tan (t //2 ) ] , y= a sin t, dy / dx=`

Options

A. ` tan t `
B. ` cos t`
C. `sec t`
D. cosec t

Answer

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

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

### Step 2: Differentiate \(x\) with respect to \(t\)
Given:
\[ x = a \left( \cos t + \log \left( \tan \left( \frac{t}{2} \right) \right) \right) \]
Differentiating \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = a \left( -\sin t + \frac{1}{\tan \left( \frac{t}{2} \right)} \cdot \frac{d}{dt} \left( \tan \left( \frac{t}{2} \right) \right) \right) \]
Using the chain rule, we know:
\[ \frac{d}{dt} \left( \tan \left( \frac{t}{2} \right) \right) = \sec^2 \left( \frac{t}{2} \right) \cdot \frac{1}{2} \]
Thus,
\[ \frac{dx}{dt} = a \left( -\sin t + \frac{1}{\tan \left( \frac{t}{2} \right)} \cdot \frac{1}{2} \sec^2 \left( \frac{t}{2} \right) \right) \]
Since \(\tan \left( \frac{t}{2} \right) = \frac{\sin \left( \frac{t}{2} \right)}{\cos \left( \frac{t}{2} \right)}\), we can rewrite:
\[ \frac{dx}{dt} = a \left( -\sin t + \frac{1}{\frac{\sin \left( \frac{t}{2} \right)}{\cos \left( \frac{t}{2} \right)}} \cdot \frac{1}{2} \cdot \frac{1}{\cos^2 \left( \frac{t}{2} \right)} \right) \]
This simplifies to:
\[ \frac{dx}{dt} = a \left( -\sin t + \frac{\cos \left( \frac{t}{2} \right)}{2 \sin \left( \frac{t}{2} \right) \cos^2 \left( \frac{t}{2} \right)} \right) \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the formula:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{a \cos t}{a \left( -\sin t + \frac{\cos \left( \frac{t}{2} \right)}{2 \sin \left( \frac{t}{2} \right) \cos^2 \left( \frac{t}{2} \right)} \right)} \]
The \(a\) cancels out:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{\cos \left( \frac{t}{2} \right)}{2 \sin \left( \frac{t}{2} \right) \cos^2 \left( \frac{t}{2} \right)}} \]

### Final Result
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos t}{-\sin t + \frac{\cos \left( \frac{t}{2} \right)}{2 \sin \left( \frac{t}{2} \right) \cos^2 \left( \frac{t}{2} \right)}} \]