It is given that `A=(tan^(-1)x)^(3)+(cot^(-1)x)^(3)` where `x gt 0` and `B=(cos^(-1)t)^(2)+(sin^(-1)t)^(2)` where `t in [0, (1)/(sqrt(2))]`, and `sin^(-1)x+cos^(-1)x=(pi)/(2)` for `-1 le x le 1` and `tan^(-1)x +cot^(-1)x=(pi)/(2)` for `x in R`.
The maximum value of B is :

Options

A. `(pi^(2))/(8)`
B. `(pi^(2))/(16)`
C. `(pi^(2))/(4)`
D. None of these

Answer

To find the maximum value of \( B = (\cos^{-1} t)^2 + (\sin^{-1} t)^2 \) where \( t \in [0, \frac{1}{\sqrt{2}}] \), we can use the identity \( \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \).

### Step-by-step Solution:

1. **Use the identity**:
\[ \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \]
Let's denote \( \sin^{-1} t = y \). Then, \( \cos^{-1} t = \frac{\pi}{2} - y \).

2. **Substitute into B**:
\[ B = \left(\cos^{-1} t\right)^2 + \left(\sin^{-1} t\right)^2 = \left(\frac{\pi}{2} - y\right)^2 + y^2 \]

3. **Expand the expression**:
\[ B = \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac{\pi}{2} \cdot y + y^2 + y^2 = \frac{\pi^2}{4} - \pi y + 2y^2 \]

4. **Differentiate B with respect to y**:
\[ \frac{dB}{dy} = -\pi + 4y \]

5. **Set the derivative to zero to find critical points**:
\[ -\pi + 4y = 0 \implies y = \frac{\pi}{4} \]

6. **Find the corresponding value of \( t \)**:
Since \( y = \sin^{-1} t \),
\[ t = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

7. **Evaluate B at the critical point**:
\[ B = \left(\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)^2 + \left(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)^2 \]
Both \( \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \) and \( \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \) equal \( \frac{\pi}{4} \):
\[ B = \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^2 = 2 \cdot \left(\frac{\pi}{4}\right)^2 = 2 \cdot \frac{\pi^2}{16} = \frac{\pi^2}{8} \]

8. **Check endpoints**:
- For \( t = 0 \):
\[ B = (\cos^{-1}(0))^2 + (\sin^{-1}(0))^2 = \left(\frac{\pi}{2}\right)^2 + 0^2 = \frac{\pi^2}{4} \]
- For \( t = \frac{1}{\sqrt{2}} \):
\[ B = \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{8} \]

9. **Conclusion**:
The maximum value of \( B \) occurs at \( t = 0 \):
\[ \text{Maximum value of } B = \frac{\pi^2}{4} \]