The value of `sin^(-1) x+cos^(-1)x(absx le 1)` is

Options

A. 1
B. `pi`
C. `pi//2`
D. `-pi//2`

Answer

To find the value of \( \sin^{-1} x + \cos^{-1} x \) given that \( |x| \leq 1 \), we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
Let \( \alpha = \sin^{-1} x \) and \( \beta = \cos^{-1} x \).

2. **Express x in terms of α and β**:
From the definitions, we have:
\[ x = \sin \alpha \quad \text{(1)} \]
\[ x = \cos \beta \quad \text{(2)} \]

3. **Set the Equations Equal**:
Since both equations (1) and (2) equal \( x \), we can set them equal to each other:
\[ \sin \alpha = \cos \beta \]

4. **Use the Co-function Identity**:
We know from trigonometric identities that:
\[ \cos \beta = \sin\left(\frac{\pi}{2} - \beta\right) \]
Therefore, we can rewrite the equation as:
\[ \sin \alpha = \sin\left(\frac{\pi}{2} - \beta\right) \]

5. **Equate Angles**:
Since the sine function is equal, we can equate the angles:
\[ \alpha = \frac{\pi}{2} - \beta \]

6. **Add α and β**:
Now, adding \( \alpha \) and \( \beta \):
\[ \alpha + \beta = \frac{\pi}{2} \]

7. **Substitute Back**:
Recall that \( \alpha = \sin^{-1} x \) and \( \beta = \cos^{-1} x \):
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]

### Final Result:
Thus, the value of \( \sin^{-1} x + \cos^{-1} x \) is:
\[ \boxed{\frac{\pi}{2}} \]