If `[sin ^(-1)(cos^(-1)(sin^(-1)(tan^( -1)x)))]=1` where [.] denotes integer function, then complete set of values of x is :

Options

A. `[tan(sin(cos1)), tan(cos(sin1))]`
B. `[tan(sin(cos1)), tan(sin(cos(sin1)))]`
C. `[tan(cos(sin1)), tan(sin(cos(sin1)))]`
D. `[tan(sin(cos1)), 1]`

Answer

To solve the equation \([ \sin^{-1}(\cos^{-1}(\sin^{-1}(\tan^{-1} x))) ] = 1\), where \([.]\) denotes the integer function, we will break it down step by step.

### Step 1: Understand the Equation
We start with the equation:
\[ \sin^{-1}(\cos^{-1}(\sin^{-1}(\tan^{-1} x))) = y \]
where \(y\) is such that \([y] = 1\). This means \(1 \leq y < 2\).

### Step 2: Analyze the Range of \(\sin^{-1}\)
The function \(\sin^{-1}(z)\) has a range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\). For \(\sin^{-1}(\tan^{-1} x)\), we need to find the range of \(\tan^{-1} x\).

### Step 3: Range of \(\tan^{-1} x\)
The range of \(\tan^{-1} x\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\). Therefore, \(\sin^{-1}(\tan^{-1} x)\) will also be within the range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

### Step 4: Substitute and Simplify
Let:
\[ z = \tan^{-1} x \implies \sin^{-1}(z) = \theta \]
Then:
\[ \sin \theta = \tan^{-1} x \]
Now we need to evaluate \(\cos^{-1}(\theta)\).

### Step 5: Evaluate \(\cos^{-1}(\theta)\)
Since \(\theta = \sin^{-1}(\tan^{-1} x)\), we have:
\[ \cos^{-1}(\theta) = \frac{\pi}{2} - \theta \]
Thus:
\[ \sin^{-1}(\cos^{-1}(\theta)) = \sin^{-1}\left(\frac{\pi}{2} - \sin^{-1}(\tan^{-1} x)\right) \]

### Step 6: Set Up the Final Equation
Now we have:
\[ \sin^{-1}\left(\frac{\pi}{2} - \sin^{-1}(\tan^{-1} x)\right) = y \]
For \([y] = 1\), we need:
\[ 1 \leq \frac{\pi}{2} - \sin^{-1}(\tan^{-1} x) < 2 \]

### Step 7: Solve the Inequalities
From \(1 \leq \frac{\pi}{2} - \sin^{-1}(\tan^{-1} x)\):
\[ \sin^{-1}(\tan^{-1} x) \leq \frac{\pi}{2} - 1 \]
And from \(\frac{\pi}{2} - \sin^{-1}(\tan^{-1} x) < 2\):
\[ \sin^{-1}(\tan^{-1} x) > \frac{\pi}{2} - 2 \]

### Step 8: Find Values of \(x\)
Now we need to find the values of \(x\) that satisfy these inequalities. We can convert \(\sin^{-1}\) back to \(x\) using the properties of inverse functions.

### Conclusion
The complete set of values of \(x\) that satisfies the original equation is:
\[ x \in \left( \tan\left(\sin\left(\frac{\pi}{2} - 1\right)\right), \tan\left(\sin\left(\frac{\pi}{2} - 2\right)\right) \right) \]