The interval of x for which `cos^(-1)x le sin^(-1)x` is :

Options

A. `(-oo,1) `
B. `[ -1, (1)/(sqrt(2)))`
C. `[(1)/(sqrt(2)),1]`
D. `[-1,1]`

Explanation

To solve the inequality \( \cos^{-1} x \leq \sin^{-1} x \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \cos^{-1} x \leq \sin^{-1} x \]
Using the identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \), we can rewrite the inequality:
\[ \cos^{-1} x \leq \frac{\pi}{2} - \cos^{-1} x \]

### Step 2: Rearrange the Inequality
Next, we rearrange the inequality:
\[ \cos^{-1} x + \cos^{-1} x \leq \frac{\pi}{2} \]
This simplifies to:
\[ 2 \cos^{-1} x \leq \frac{\pi}{2} \]

### Step 3: Divide by 2
Now, we divide both sides by 2:
\[ \cos^{-1} x \leq \frac{\pi}{4} \]

### Step 4: Apply the Cosine Function
To find the corresponding values of \( x \), we apply the cosine function to both sides. Remember that the cosine function is decreasing in the interval \([0, \pi]\):
\[ x \geq \cos\left(\frac{\pi}{4}\right) \]
Since \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), we have:
\[ x \geq \frac{1}{\sqrt{2}} \]

### Step 5: Consider the Domain of \( \cos^{-1} x \)
The domain of \( \cos^{-1} x \) is \( x \in [-1, 1] \). Therefore, we need to combine this with our previous result:
\[ \frac{1}{\sqrt{2}} \leq x \leq 1 \]

### Final Interval
Thus, the interval of \( x \) for which \( \cos^{-1} x \leq \sin^{-1} x \) is:
\[ \left[\frac{1}{\sqrt{2}}, 1\right] \]