Let [.] represents the greatest integer function and `[cos^(-1)sin^(-1)tan^(-1)x]=1` then 'x' lies in the intervel.

Options

A. `[sintancos2,tansincos1]`
B. `[tansincos2,tansincos1]`
C. `[tansincos1, tan sincos2]`
D. `(tansincos2,tansincos1]`

Answer

To solve the problem, we need to analyze the expression \([ \cos^{-1}(\sin^{-1}(\tan^{-1}(x))) ] = 1\) and determine the interval in which \(x\) lies.

### Step-by-Step Solution:

1. **Understanding the Greatest Integer Function**:
The greatest integer function \([y]\) gives the largest integer less than or equal to \(y\). For \([y] = 1\), it implies that \(1 \leq y < 2\).

2. **Setting Up the Inequality**:
From the problem, we have:
\[ 1 \leq \cos^{-1}(\sin^{-1}(\tan^{-1}(x))) < 2 \]

3. **Applying the Cosine Function**:
Since \(\cos^{-1}(y)\) is a decreasing function, we can apply the cosine to all parts of the inequality:
\[ \cos(2) < \sin^{-1}(\tan^{-1}(x)) \leq \cos(1) \]

4. **Understanding the Sine Function**:
The sine function is increasing, so we can apply the sine to the entire inequality:
\[ \sin(\cos(2)) < \tan^{-1}(x) \leq \sin(\cos(1)) \]

5. **Applying the Tangent Function**:
The tangent function is also increasing, so we can apply the tangent to the entire inequality:
\[ \tan(\sin(\cos(2))) < x \leq \tan(\sin(\cos(1))) \]

6. **Finding the Interval for \(x\)**:
Now we need to compute the values of \(\tan(\sin(\cos(2)))\) and \(\tan(\sin(\cos(1)))\) to find the interval for \(x\).

7. **Calculating Values**:
- Compute \(\cos(2)\) and \(\cos(1)\).
- Compute \(\sin(\cos(2))\) and \(\sin(\cos(1))\).
- Finally, compute \(\tan(\sin(\cos(2)))\) and \(\tan(\sin(\cos(1)))\).

8. **Conclusion**:
The interval for \(x\) will be:
\[ \tan(\sin(\cos(2))) < x \leq \tan(\sin(\cos(1))) \]