The soluation set of inequality `(sinx+cos^(- 1)x)-(cosx-sin^-1x)>= pi/2,` is equal to

Options

A. `[(pi)/(4),(5pi)/(4)]`
B. `uuu_(n in I)[2n pi+(pi)/(4),2n pi+(5pi)/(4)]`
C. `[(pi)/(4),1]`
D. `[-1,(-pi)/(4)]uu[(pi)/(4),1]`

Explanation

To solve the inequality \((\sin x + \cos^{-1} x) - (\cos x - \sin^{-1} x) \geq \frac{\pi}{2}\), we will follow these steps:

### Step 1: Rewrite the Inequality
Start by rewriting the inequality:
\[ \sin x + \cos^{-1} x - \cos x + \sin^{-1} x \geq \frac{\pi}{2} \]

### Step 2: Use the Identity for Inverse Trigonometric Functions
Recall the identity:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Using this identity, we can substitute \(\sin^{-1} x + \cos^{-1} x\) in the inequality:
\[ \sin x + \left(\frac{\pi}{2} - \sin^{-1} x\right) - \cos x + \sin^{-1} x \geq \frac{\pi}{2} \]

### Step 3: Simplify the Expression
Now simplify the expression:
\[ \sin x - \cos x + \frac{\pi}{2} \geq \frac{\pi}{2} \]
Subtract \(\frac{\pi}{2}\) from both sides:
\[ \sin x - \cos x \geq 0 \]

### Step 4: Rearranging the Inequality
This can be rearranged to:
\[ \sin x \geq \cos x \]

### Step 5: Determine the Solution Set
The inequality \(\sin x \geq \cos x\) holds true in the interval where:
\[ \tan x \geq 1 \]
This occurs in the intervals:
\[ x \in \left[\frac{\pi}{4} + n\pi, \frac{5\pi}{4} + n\pi\right] \quad \text{for } n \in \mathbb{Z} \]

### Step 6: Consider the Domain
Since we are dealing with the functions \(\sin^{-1} x\) and \(\cos^{-1} x\), the domain of \(x\) is restricted to:
\[ x \in [-1, 1] \]

### Step 7: Find the Intersection of the Domain and Solution Set
Now we need to find the intersection of the intervals from the inequality with the domain:
- The relevant interval for \(\sin x \geq \cos x\) in the domain \([-1, 1]\) is:
\[ x \in \left[\frac{\pi}{4}, 1\right] \]

### Final Solution
Thus, the solution set of the inequality is:
\[ \left[\frac{\pi}{4}, 1\right] \]