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Let's break down the problem step by step. It seems to be focused on the geometry and calculus of an object moving around a unit circle, and we need to deal with the tangent line and the area of a triangle formed.

### Part (a)
The slope of the tangent line to the unit circle at \( P(t) \) is given as:
\[
\text{slope} = -\cot(t)
\]
This is correct, as the derivative of the parametric equation for the unit circle (where \( x = \cos(t) \) and \( y = \sin(t) \)) gives \( \frac{dy}{dx} = -\cot(t) \).

### Part (b)
The point-slope form for the equation of the tangent line is:
\[
y - \sin(t) = -\cot(t) \cdot (x - \cos(t))
\]
Again, this is correct. The equation is derived by plugging in the point \( P(t) = (\cos(t), \sin(t)) \) and the slope \( -\cot(t) \) into the point-slope form of the equation of a line.

### Part (c)
You are asked to find the area of the right triangle formed by the tangent line, the x-axis, and the y-axis. The base and height of this triangle can be found as follows:

- The x-intercept occurs when \( y = 0 \) in the tangent line equation. Solving \( 0 - \sin(t) = -\cot(t)(x - \cos(t)) \), the x-intercept is:
  \[
  x = \frac{\sin(t)}{\cot(t)} + \cos(t) = \csc(t)
  \]
- The y-intercept occurs when \( x = 0 \) in the tangent line equation. Plugging into the line equation gives:
  \[
  y - \sin(t) = -\cot(t)(0 - \cos(t)) \implies y = \cot(t) \cos(t) + \sin(t) = \csc(t)
  \]
  
Since the triangle has both the base and height equal to \( \csc(t) \), the area of the triangle is:
\[
a(t) = \frac{1}{2} \times \csc(t) \times \csc(t) = \frac{1}{2} \csc^2(t)
\]

### Part (d)
Now, you are asked to compute the limit of \( a(t) \) as \( t \to \frac{\pi}{2} \). Recall that \( \csc(t) = \frac{1}{\sin(t)} \). As \( t \to \frac{\pi}{2} \), \( \sin(t) \to 1 \), so \( \csc(t) \to 1 \). Therefore:
\[
\lim_{t \to \frac{\pi}{2}} a(t) = \frac{1}{2} \times 1^2 = \frac{1}{2}
\]

### Part (e)
Finally, you are asked to compute the limit of \( a(t) \) as \( t \to 0^+ \). As \( t \to 0 \), \( \sin(t) \to 0 \), so \( \csc(t) \to \infty \). Thus:
\[
\lim_{t \to 0^+} a(t) = \lim_{t \to 0^+} \frac{1}{2} \csc^2(t) = \frac{1}{2} \times \infty = \infty
\]

### Summary:
1. \( \text{slope of tangent line} = -\cot(t) \)
2. \( \text{Equation of tangent line} = y - \sin(t) = -\cot(t)(x - \cos(t)) \)
3. \( a(t) = \frac{1}{2} \csc^2(t) \)
4. \( \lim_{t \to \frac{\pi}{2}} a(t) = \frac{1}{2} \)
5. \( \lim_{t \to 0^+} a(t) = \infty \)

Would you like to go into more detail for any specific part?

Here are 5 related questions:
1. What is the geometric interpretation of the cotangent function in relation to the unit circle?
2. How is the equation of a tangent line derived from parametric equations?
3. Why does \( a(t) \) approach infinity as \( t \to 0^+ \)?
4. What would happen to the triangle if \( t \) were greater than \( \frac{\pi}{2} \)?
5. How do limits help in understanding the behavior of a function near critical points?

**Tip:** Always double-check the geometry when working with parametric equations and tangent lines—this can help ensure that you interpret the results correctly.

An object is moving around the unit circle, and at a given time t, the tangent line to the unit circle at the position P(t) forms a right triangle. Using the parametric equations x = cos(t) and y = sin(t), find the slope of the tangent line, the equation of the tangent line, the area of the triangle, and limits of the area as t approaches critical values.

Solve the problem of finding the slope of the tangent line and the area of a right triangle formed by a tangent to the unit circle. Learn about parametric equations, the cotangent function, and how limits apply to the triangle's area as t approaches π/2 and 0.

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