Is ` sec^(-1) (-x) = pi - sec^(-1) x , |x| ge 1 `?

Answer

To prove the statement \( \sec^{-1}(-x) = \pi - \sec^{-1}(x) \) for \( |x| \geq 1 \), we will follow these steps:

### Step 1: Understanding the Definition of \( \sec^{-1}(x) \)
The function \( \sec^{-1}(x) \) is defined for \( |x| \geq 1 \). The range of \( \sec^{-1}(x) \) is \( [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] \).

### Step 2: Analyzing \( \sec^{-1}(-x) \)
For \( x \geq 1 \), we have \( -x \leq -1 \). Thus, \( \sec^{-1}(-x) \) is defined and we need to find its value.

### Step 3: Using the Property of the Secant Function
Recall that:
\[ \sec(\theta) = \frac{1}{\cos(\theta)} \]
If \( \sec^{-1}(x) = \theta \), then \( \sec(\theta) = x \). For \( \sec^{-1}(-x) \), we have:
\[ \sec(\phi) = -x \implies \phi = \sec^{-1}(-x) \]

### Step 4: Relating Angles
From the properties of the secant function, we know:
\[ \sec(\pi - \theta) = -\sec(\theta) \]
This implies:
\[ \sec(\phi) = -x \implies \phi = \pi - \theta \]
where \( \theta = \sec^{-1}(x) \).

### Step 5: Conclusion
Thus, we can conclude:
\[ \sec^{-1}(-x) = \pi - \sec^{-1}(x) \]
for \( |x| \geq 1 \).

### Final Statement
Therefore, the statement \( \sec^{-1}(-x) = \pi - \sec^{-1}(x) \) is true for \( |x| \geq 1 \).

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