Statement I `sin ^(-1)(1/sqrte) gt tan^(-1)(1/sqrtpi)`
Statement II `sin^(-1) x gt tan^(-1) y " for " x gt y , AA x, y in (0,1)`

Options

A. Statement I is True, Statement II is True, Statement II is a correct explanation for statement I
B. Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement I
C. Statement I is True, Statement II is False
D. Statement I is False, Statement II is True.

Answer

To solve the given problem, we will analyze both statements step by step.

### Statement I:
We need to verify if \( \sin^{-1}\left(\frac{1}{\sqrt{e}}\right) > \tan^{-1}\left(\frac{1}{\sqrt{\pi}}\right) \).

1. **Calculate \( \sin^{-1}\left(\frac{1}{\sqrt{e}}\right) \)**:
- We know that \( e \approx 2.718 \), so \( \sqrt{e} \approx \sqrt{2.718} \approx 1.6487 \).
- Therefore, \( \frac{1}{\sqrt{e}} \approx \frac{1}{1.6487} \approx 0.607 \).
- Now, we find \( \sin^{-1}(0.607) \). The value of \( \sin^{-1}(0.607) \) is approximately \( 0.6435 \) radians (or about \( 36.87^\circ \)).

2. **Calculate \( \tan^{-1}\left(\frac{1}{\sqrt{\pi}}\right) \)**:
- We know that \( \pi \approx 3.14 \), so \( \sqrt{\pi} \approx \sqrt{3.14} \approx 1.772 \).
- Therefore, \( \frac{1}{\sqrt{\pi}} \approx \frac{1}{1.772} \approx 0.564 \).
- Now, we find \( \tan^{-1}(0.564) \). The value of \( \tan^{-1}(0.564) \) is approximately \( 0.515 \) radians (or about \( 29.74^\circ \)).

3. **Compare the two values**:
- We have \( \sin^{-1}\left(\frac{1}{\sqrt{e}}\right) \approx 0.6435 \) and \( \tan^{-1}\left(\frac{1}{\sqrt{\pi}}\right) \approx 0.515 \).
- Since \( 0.6435 > 0.515 \), we conclude that \( \sin^{-1}\left(\frac{1}{\sqrt{e}}\right) > \tan^{-1}\left(\frac{1}{\sqrt{\pi}}\right) \).

Thus, **Statement I is true**.

### Statement II:
We need to verify if \( \sin^{-1}(x) > \tan^{-1}(y) \) for \( x > y \) where \( x, y \in (0, 1) \).

1. **Understanding the functions**:
- Both \( \sin^{-1}(x) \) and \( \tan^{-1}(y) \) are increasing functions in the interval \( (0, 1) \).
- This means that if \( x_1 > x_2 \), then \( \sin^{-1}(x_1) > \sin^{-1}(x_2) \) and similarly for \( \tan^{-1} \).

2. **Applying the property of increasing functions**:
- Given \( x > y \), since both functions are increasing, we can conclude that \( \sin^{-1}(x) > \sin^{-1}(y) \) and \( \tan^{-1}(x) > \tan^{-1}(y) \).

3. **Conclusion**:
- Therefore, if \( x > y \), it follows that \( \sin^{-1}(x) > \tan^{-1}(y) \) holds true.

Thus, **Statement II is also true**.

### Final Conclusion:
Both statements are true.

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