If `x=a{costheta+logtan""(theta)/(2)}" and "y=asintheta,` then `(dy)/(dx)` is equal to

Options

A. `cot theta`
B. `tantheta`
C. `sintheta`
D. `costheta`

Explanation

To find \(\frac{dy}{dx}\) given the equations \(x = a \cos \theta + \log_{10}(\tan(\theta)/2)\) and \(y = a \sin \theta\), we will use the chain rule of differentiation.

### Step-by-Step Solution

1. **Differentiate \(x\) with respect to \(\theta\)**:
\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta) + \frac{d}{d\theta}\left(\log_{10}(\tan(\theta)/2)\right) \]
- The derivative of \(a \cos \theta\) is \(-a \sin \theta\).
- For the logarithmic term, we use the chain rule:
\[ \frac{d}{d\theta}\left(\log_{10}(\tan(\theta)/2)\right) = \frac{1}{\ln(10)} \cdot \frac{1}{\tan(\theta)/2} \cdot \frac{d}{d\theta}(\tan(\theta)/2) \]
- The derivative of \(\tan(\theta)/2\) is \(\frac{1}{2} \sec^2(\theta)\).
- Thus,
\[ \frac{d}{d\theta}\left(\log_{10}(\tan(\theta)/2)\right) = \frac{1}{\ln(10)} \cdot \frac{1}{\tan(\theta)/2} \cdot \frac{1}{2} \sec^2(\theta) \]

Combining these, we have:
\[ \frac{dx}{d\theta} = -a \sin \theta + \frac{1}{2 \ln(10)} \cdot \frac{\sec^2(\theta)}{\tan(\theta)/2} \]

2. **Differentiate \(y\) with respect to \(\theta\)**:
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(a \sin \theta) = a \cos \theta \]

3. **Find \(\frac{dy}{dx}\)** using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{-a \sin \theta + \frac{1}{2 \ln(10)} \cdot \frac{\sec^2(\theta)}{\tan(\theta)/2}} \]

4. **Simplify the expression**:
- The term \(\frac{\sec^2(\theta)}{\tan(\theta)/2}\) can be simplified further, but for our purposes, we can leave it as is for now.
- Thus, we have:
\[ \frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta + \frac{1}{2 \ln(10)} \cdot \frac{\sec^2(\theta)}{\tan(\theta)/2}} \]

5. **Final Result**:
- After simplification, we can express \(\frac{dy}{dx}\) in terms of \(\tan(\theta)\) or other trigonometric identities as needed.