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 the condition \( 3 \leq a < 4 \).

### Step-by-Step Solution:

1. **Identify the Greatest Integer Function**:
Since \( a \) is in the range \( 3 \leq a < 4 \), the greatest integer function \( [a] \) is equal to 3. Thus, we can rewrite the expression as:
\[ \sin^{-1}(\sin[3]) + \tan^{-1}(\tan[3]) + \sec^{-1}(\sec[3]) \]

2. **Evaluate \( \sin^{-1}(\sin[3]) \)**:
The value of \( 3 \) is in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \). For values of \( x \) in this interval, the property of the sine inverse function gives us:
\[ \sin^{-1}(\sin x) = \pi - x \]
Therefore:
\[ \sin^{-1}(\sin[3]) = \pi - 3 \]

3. **Evaluate \( \tan^{-1}(\tan[3]) \)**:
Similarly, since \( 3 \) is also in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), we use the property of the tangent inverse function:
\[ \tan^{-1}(\tan x) = x - \pi \]
Thus:
\[ \tan^{-1}(\tan[3]) = 3 - \pi \]

4. **Evaluate \( \sec^{-1}(\sec[3]) \)**:
The value of \( 3 \) is in the interval \( (0, \pi) \), where the secant inverse function behaves as:
\[ \sec^{-1}(\sec x) = x \]
Therefore:
\[ \sec^{-1}(\sec[3]) = 3 \]

5. **Combine All Parts**:
Now we can combine all the evaluated parts:
\[ \sin^{-1}(\sin[3]) + \tan^{-1}(\tan[3]) + \sec^{-1}(\sec[3]) = (\pi - 3) + (3 - \pi) + 3 \]
Simplifying this gives:
\[ = \pi - 3 + 3 - \pi + 3 = 3 \]

### Final Answer:
Thus, the value of the expression is:
\[ \boxed{3} \]