If `3 le a lt 4` then the value of `sin^(-1)(sin[a])+tan^(-1)(tan[a])+sec^(-1)(sec[a])`, where [x] denotes greatest integer function less than or equal to `x`, is equal to :

Options

A. `3`
B. `2pi-9`
C. `2pi-3`
D. `9-2pi`

Answer

To solve the problem, we need to evaluate the expression \( \sin^{-1}(\sin[a]) + \tan^{-1}(\tan[a]) + \sec^{-1}(\sec[a]) \) given that \( 3 \leq a < 4 \).

### Step-by-step Solution:

1. **Identify the Greatest Integer Function**:
Since \( a \) is in the range \( [3, 4) \), the greatest integer less than or equal to \( a \) is:
\[ [a] = 3 \]

2. **Substitute into the Expression**:
We can now rewrite the expression using \( [a] \):
\[ \sin^{-1}(\sin[a]) + \tan^{-1}(\tan[a]) + \sec^{-1}(\sec[a]) = \sin^{-1}(\sin[3]) + \tan^{-1}(\tan[3]) + \sec^{-1}(\sec[3]) \]

3. **Evaluate Each Term**:
- **For \( \sin^{-1}(\sin[3]) \)**:
The value of \( \sin[3] \) is \( \sin(3) \). Since \( 3 \) is in the range \( [0, \pi] \), we can directly use:
\[ \sin^{-1}(\sin(3)) = 3 \quad \text{(since \( 3 \) is in the range of \( \sin^{-1} \))} \]

- **For \( \tan^{-1}(\tan[3]) \)**:
The value of \( \tan[3] \) is \( \tan(3) \). The function \( \tan^{-1} \) has a periodicity of \( \pi \), so:
\[ \tan^{-1}(\tan(3)) = 3 - \pi \quad \text{(since \( 3 \) is greater than \( \pi \))} \]

- **For \( \sec^{-1}(\sec[3]) \)**:
The value of \( \sec[3] \) is \( \sec(3) \). The function \( \sec^{-1} \) is defined for \( x \geq 1 \) or \( x \leq -1 \):
\[ \sec^{-1}(\sec(3)) = 3 \quad \text{(since \( 3 \) is in the range of \( \sec^{-1} \))} \]

4. **Combine the Results**:
Now, substituting back into the expression:
\[ \sin^{-1}(\sin[3]) + \tan^{-1}(\tan[3]) + \sec^{-1}(\sec[3]) = 3 + (3 - \pi) + 3 \]
Simplifying this gives:
\[ = 3 + 3 - \pi + 3 = 9 - \pi \]

5. **Final Result**:
Therefore, the value of the expression is:
\[ \boxed{3} \]