For `x, y, z, t in R, sin^(-1) x + cos^(-1) y + sec^(-1) z ge t^(2) - sqrt(2pi t) + 3pi`
The principal value of `cos^(-1) (cos 5t^(2))` is

Options

A. `(3pi)/(2)`
B. `(pi)/(2)`
C. `(pi)/(3)`
D. `(2pi)/(3)`

Explanation

To solve the problem step-by-step, we will analyze the given inequality and find the principal value of \( \cos^{-1}(\cos(5t^2)) \).

### Step 1: Understand the inequality
We start with the inequality:
\[ \sin^{-1}(x) + \cos^{-1}(y) + \sec^{-1}(z) \geq t^2 - \sqrt{2\pi t} + 3\pi \]

### Step 2: Analyze the ranges of the inverse functions
The ranges of the inverse trigonometric functions are:
- \( \sin^{-1}(x) \) is in \([- \frac{\pi}{2}, \frac{\pi}{2}]\)
- \( \cos^{-1}(y) \) is in \([0, \pi]\)
- \( \sec^{-1}(z) \) is in \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\)

### Step 3: Find the maximum value of the left-hand side
The maximum value of \( \sin^{-1}(x) + \cos^{-1}(y) + \sec^{-1}(z) \) occurs when:
- \( \sin^{-1}(x) = \frac{\pi}{2} \) (which implies \( x = 1 \))
- \( \cos^{-1}(y) = \pi \) (which implies \( y = -1 \))
- \( \sec^{-1}(z) = \frac{\pi}{2} \) (which implies \( z = 1 \))

Thus, the maximum value of the left-hand side is:
\[ \frac{\pi}{2} + \pi + \frac{\pi}{2} = 2\pi \]

### Step 4: Analyze the right-hand side
Now, we need to analyze the right-hand side:
\[ t^2 - \sqrt{2\pi t} + 3\pi \]
To find the minimum value of this expression, we can complete the square.

### Step 5: Completing the square
The expression can be rewritten as:
\[ t^2 - \sqrt{2\pi}t + 3\pi \]
Completing the square:
\[ = \left(t - \frac{\sqrt{2\pi}}{2}\right)^2 + \left(3\pi - \frac{2\pi}{4}\right) \]
This shows that the minimum value occurs at \( t = \frac{\sqrt{2\pi}}{2} \).

### Step 6: Find the minimum value
Substituting \( t = \frac{\sqrt{2\pi}}{2} \):
\[ \left(\frac{\sqrt{2\pi}}{2}\right)^2 - \sqrt{2\pi}\left(\frac{\sqrt{2\pi}}{2}\right) + 3\pi = \frac{2\pi}{4} - \frac{2\pi}{2} + 3\pi = -\frac{2\pi}{4} + 3\pi = \frac{10\pi}{4} = \frac{5\pi}{2} \]

### Step 7: Set the inequality
Thus, we have:
\[ \sin^{-1}(1) + \cos^{-1}(-1) + \sec^{-1}(1) \geq \frac{5\pi}{2} \]
This implies:
\[ \frac{\pi}{2} + \pi + \frac{\pi}{2} = 2\pi \geq \frac{5\pi}{2} \]
This inequality holds true.

### Step 8: Find the principal value of \( \cos^{-1}(\cos(5t^2)) \)
Now, we need to find \( \cos^{-1}(\cos(5t^2)) \):
\[ t = \frac{\sqrt{2\pi}}{2} \implies t^2 = \frac{2\pi}{4} = \frac{\pi}{2} \]
Thus,
\[ 5t^2 = 5 \cdot \frac{\pi}{2} = \frac{5\pi}{2} \]

### Step 9: Calculate \( \cos^{-1}(\cos(\frac{5\pi}{2})) \)
The angle \( \frac{5\pi}{2} \) can be simplified:
\[ \frac{5\pi}{2} - 2\pi = \frac{\pi}{2} \]
Thus,
\[ \cos^{-1}(\cos(\frac{5\pi}{2})) = \cos^{-1}(\cos(\frac{\pi}{2})) = \frac{\pi}{2} \]

### Final Answer
The principal value of \( \cos^{-1}(\cos(5t^2)) \) is:
\[ \frac{\pi}{2} \]