`sin^(-1)x+cos^(-1)x` is equal to

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `-1`
D. 1

Answer

To solve the problem \( \sin^{-1}x + \cos^{-1}x \), we can follow these steps:

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

### Step 2: Use the Definitions
From the definitions, we have:
- \( \sin \alpha = x \)
- \( \cos \beta = x \)

### Step 3: Use the Pythagorean Identity
We know that for any angle \( \alpha \):
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
Since \( \beta = \cos^{-1}x \), we can also express \( \cos \beta \) in terms of \( \alpha \):
\[ \cos \beta = \sin\left(\frac{\pi}{2} - \beta\right) = \sin \alpha \]
Thus, we can write:
\[ \sin \alpha = \cos \beta \]

### Step 4: Relate \( \alpha \) and \( \beta \)
From the relationship between sine and cosine, we can express \( \beta \) in terms of \( \alpha \):
\[ \sin \alpha = \cos \left(\frac{\pi}{2} - \alpha\right) \]
This means:
\[ \beta = \frac{\pi}{2} - \alpha \]

### Step 5: Add the Angles
Now, we can add \( \alpha \) and \( \beta \):
\[ \alpha + \beta = \alpha + \left(\frac{\pi}{2} - \alpha\right) = \frac{\pi}{2} \]

### Conclusion
Thus, we conclude that:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]

### Final Answer
The value of \( \sin^{-1}x + \cos^{-1}x \) is \( \frac{\pi}{2} \).

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