Find the slopes of the tangent and the normal to the curve `y=(sin2x+cotx+2)^2` at `x=pi//2`

Explanation

To find the slopes of the tangent and the normal to the curve \( y = (\sin 2x + \cot x + 2)^2 \) at \( x = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative \( \frac{dy}{dx} \) using the chain rule.

Given:
\[ y = (\sin 2x + \cot x + 2)^2 \]

Using the chain rule:
\[ \frac{dy}{dx} = 2(\sin 2x + \cot x + 2) \cdot \frac{d}{dx}(\sin 2x + \cot x + 2) \]

### Step 2: Differentiate the inner function
Now we differentiate the inner function \( \sin 2x + \cot x + 2 \).

The derivatives are:
- \( \frac{d}{dx}(\sin 2x) = 2\cos 2x \)
- \( \frac{d}{dx}(\cot x) = -\csc^2 x \)
- The derivative of the constant \( 2 \) is \( 0 \).

Thus,
\[ \frac{d}{dx}(\sin 2x + \cot x + 2) = 2\cos 2x - \csc^2 x \]

### Step 3: Substitute back into the derivative
Substituting back, we have:
\[ \frac{dy}{dx} = 2(\sin 2x + \cot x + 2)(2\cos 2x - \csc^2 x) \]

### Step 4: Evaluate the derivative at \( x = \frac{\pi}{2} \)
Now we need to evaluate \( \frac{dy}{dx} \) at \( x = \frac{\pi}{2} \).

First, calculate \( \sin 2\left(\frac{\pi}{2}\right) \):
\[ \sin(\pi) = 0 \]

Next, calculate \( \cot\left(\frac{\pi}{2}\right) \):
\[ \cot\left(\frac{\pi}{2}\right) = 0 \]

Thus,
\[ \sin 2x + \cot x + 2 = 0 + 0 + 2 = 2 \]

Now calculate \( \cos 2\left(\frac{\pi}{2}\right) \):
\[ \cos(\pi) = -1 \]

And calculate \( \csc^2\left(\frac{\pi}{2}\right) \):
\[ \csc\left(\frac{\pi}{2}\right) = 1 \implies \csc^2\left(\frac{\pi}{2}\right) = 1 \]

Substituting these values into the derivative:
\[ \frac{dy}{dx} = 2(2)(2(-1) - 1) = 2(2)(-2 - 1) = 2(2)(-3) = -12 \]

### Step 5: Find the slope of the normal
The slope of the normal \( m_2 \) is given by:
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-12} = \frac{1}{12} \]

### Final Result
- The slope of the tangent at \( x = \frac{\pi}{2} \) is \( -12 \).
- The slope of the normal at \( x = \frac{\pi}{2} \) is \( \frac{1}{12} \).