If `alpha="sin"^(-1)("cos"("sin"^(-1)x))` and `beta="cos"^(-1)("sin"("cos"^(-1)x))`, then

Options

A. `"tan" alpha=tan beta`
B. `"tan" alpha=-cot beta`
C. `"tan" alpha="cot" beta`
D. `"tan" alpha=-"tan" beta`

Answer

To solve the problem, we need to analyze the expressions for \( \alpha \) and \( \beta \) given in the question:

1. **Given:**
\[ \alpha = \sin^{-1}(\cos(\sin^{-1} x)) \]
\[ \beta = \cos^{-1}(\sin(\cos^{-1} x)) \]

2. **Using the identity:**
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This means we can express \( \beta \) in terms of \( \alpha \).

3. **Rewriting \( \beta \):**
We can rewrite \( \beta \) as follows:
\[ \beta = \cos^{-1}(\sin(\cos^{-1} x)) = \cos^{-1}(\sin(\frac{\pi}{2} - \sin^{-1} x)) \]
Since \( \sin(\frac{\pi}{2} - \theta) = \cos(\theta) \), we can say:
\[ \beta = \cos^{-1}(\cos(\sin^{-1} x)) \]

4. **Using the identity:**
From the identity \( \cos^{-1}(\cos \theta) = \theta \) for \( \theta \) in the range \( [0, \pi] \), we find:
\[ \beta = \sin^{-1} x \]

5. **Now, we have:**
\[ \alpha + \beta = \frac{\pi}{2} \]
This implies:
\[ \alpha = \frac{\pi}{2} - \beta \]

6. **Taking the tangent:**
We can now take the tangent of both sides:
\[ \tan \alpha = \tan\left(\frac{\pi}{2} - \beta\right) \]
Using the identity \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \):
\[ \tan \alpha = \cot \beta \]

7. **Final result:**
Thus, we conclude:
\[ \tan \alpha = \cot \beta \]

### Conclusion:
The correct option is:
- **Option C:** \( \tan \alpha = \cot \beta \)