Prove that `d/(dx)(cosectheta)=-cosecthetacottheta.`

Answer

To prove that \(\frac{d}{d\theta}(\csc \theta) = -\csc \theta \cot \theta\), we will use the quotient rule of differentiation.

### Step-by-Step Solution:

1. **Rewrite the cosecant function**:
\[ \csc \theta = \frac{1}{\sin \theta} \]

2. **Apply the quotient rule**:
The quotient rule states that if \(y = \frac{u}{v}\), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, let \(u = 1\) and \(v = \sin \theta\). Therefore:
\[ \frac{d}{d\theta}(\csc \theta) = \frac{\sin \theta \cdot \frac{d}{d\theta}(1) - 1 \cdot \frac{d}{d\theta}(\sin \theta)}{\sin^2 \theta} \]

3. **Differentiate the components**:
- The derivative of a constant (1) is 0:
\[ \frac{d}{d\theta}(1) = 0 \]
- The derivative of \(\sin \theta\) is \(\cos \theta\):
\[ \frac{d}{d\theta}(\sin \theta) = \cos \theta \]

4. **Substitute the derivatives back into the quotient rule**:
\[ \frac{d}{d\theta}(\csc \theta) = \frac{\sin \theta \cdot 0 - 1 \cdot \cos \theta}{\sin^2 \theta} \]
This simplifies to:
\[ \frac{d}{d\theta}(\csc \theta) = \frac{-\cos \theta}{\sin^2 \theta} \]

5. **Rewrite using cosecant and cotangent**:
We know that:
\[ \csc \theta = \frac{1}{\sin \theta} \quad \text{and} \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Therefore:
\[ \frac{-\cos \theta}{\sin^2 \theta} = -\frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta} = -\cot \theta \cdot \csc \theta \]

6. **Final result**:
Thus, we have:
\[ \frac{d}{d\theta}(\csc \theta) = -\csc \theta \cot \theta \]