Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=ctantheta,y=c cottheta`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = c \tan \theta\) and \(y = c \cot \theta\), we will differentiate both equations with respect to \(\theta\) and then use the chain rule to find \(\frac{dy}{dx}\).

### Step-by-Step Solution:

1. **Differentiate \(x\) with respect to \(\theta\)**:
\[ x = c \tan \theta \]
The derivative of \(\tan \theta\) is \(\sec^2 \theta\). Therefore,
\[ \frac{dx}{d\theta} = c \sec^2 \theta \]

2. **Differentiate \(y\) with respect to \(\theta\)**:
\[ y = c \cot \theta \]
The derivative of \(\cot \theta\) is \(-\csc^2 \theta\). Therefore,
\[ \frac{dy}{d\theta} = -c \csc^2 \theta \]

3. **Use the chain rule to find \(\frac{dy}{dx}\)**:
According to the chain rule,
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{-c \csc^2 \theta}{c \sec^2 \theta} \]

4. **Simplify the expression**:
The \(c\) in the numerator and denominator cancels out:
\[ \frac{dy}{dx} = \frac{-\csc^2 \theta}{\sec^2 \theta} \]
We know that \(\csc \theta = \frac{1}{\sin \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\), so:
\[ \frac{dy}{dx} = -\frac{1/\sin^2 \theta}{1/\cos^2 \theta} = -\frac{\cos^2 \theta}{\sin^2 \theta} \]
This simplifies to:
\[ \frac{dy}{dx} = -\cot^2 \theta \]

### Final Answer:
\[ \frac{dy}{dx} = -\cot^2 \theta \]