If `a le sin^(-1) x+cos^(-1)x+tan^(-1) x le b`, then

Options

A. `a=0,b=pi`
B. `a=0,b=pi/2`
C. `a=pi/2,b=pi`
D. none

Answer

To solve the inequality \( a \leq \sin^{-1} x + \cos^{-1} x + \tan^{-1} x \leq b \), we can follow these steps:

### Step 1: Use the identity for \(\sin^{-1} x + \cos^{-1} x\)
We know from trigonometric identities that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Thus, we can rewrite the expression:
\[ \sin^{-1} x + \cos^{-1} x + \tan^{-1} x = \frac{\pi}{2} + \tan^{-1} x \]

### Step 2: Substitute into the inequality
Now, substituting this back into the inequality gives us:
\[ a \leq \frac{\pi}{2} + \tan^{-1} x \leq b \]

### Step 3: Isolate \(\tan^{-1} x\)
Next, we can isolate \(\tan^{-1} x\) by subtracting \(\frac{\pi}{2}\) from all parts of the inequality:
\[ a - \frac{\pi}{2} \leq \tan^{-1} x \leq b - \frac{\pi}{2} \]

### Step 4: Determine the range of \(\tan^{-1} x\)
The function \(\tan^{-1} x\) has a range of:
\[ -\frac{\pi}{2} < \tan^{-1} x < \frac{\pi}{2} \]
This means that:
\[ -\frac{\pi}{2} \leq \tan^{-1} x \leq \frac{\pi}{2} \]

### Step 5: Compare intervals
Now we can compare the intervals:
1. From \( a - \frac{\pi}{2} \leq \tan^{-1} x \), we have \( a - \frac{\pi}{2} \geq -\frac{\pi}{2} \) which simplifies to:
\[ a \geq 0 \]
2. From \( \tan^{-1} x \leq b - \frac{\pi}{2} \), we have \( \frac{\pi}{2} \leq b - \frac{\pi}{2} \) which simplifies to:
\[ b \leq \pi \]

### Conclusion
Thus, we conclude that:
\[ 0 \leq a \leq b \leq \pi \]
The values of \( a \) and \( b \) are:
\[ a = 0 \quad \text{and} \quad b = \pi \]

### Final Answer
The final answer is:
\[ a = 0, \quad b = \pi \]