If the maximum value of `(sec^-1 x)^2 + (cosec^-1 x)^2` approaches a, the minimum value of `(tan^-1 x)^3 +(cot^-1 x)^3` approaches b then `(a+b/pi)` is equal to

Options

A. `pi^2`
B. `(41pi^2)/32`
C. `pi^2/32`
D. `(43pi^2)/32`

Answer

To solve the problem, we need to find the maximum value of \((\sec^{-1} x)^2 + (\csc^{-1} x)^2\) and the minimum value of \((\tan^{-1} x)^3 + (\cot^{-1} x)^3\), and then compute \(a + \frac{b}{\pi}\).

### Step 1: Finding the maximum value of \((\sec^{-1} x)^2 + (\csc^{-1} x)^2\)

Let:
\[ y = (\sec^{-1} x)^2 + (\csc^{-1} x)^2 \]
Using the identity:
\[ \sec^{-1} x + \csc^{-1} x = \frac{\pi}{2} \]
we can express \(\csc^{-1} x\) in terms of \(\sec^{-1} x\):
\[ \csc^{-1} x = \frac{\pi}{2} - \sec^{-1} x \]
Substituting this into \(y\):
\[ y = (\sec^{-1} x)^2 + \left(\frac{\pi}{2} - \sec^{-1} x\right)^2 \]
Let \(t = \sec^{-1} x\):
\[ y = t^2 + \left(\frac{\pi}{2} - t\right)^2 \]
Expanding the square:
\[ y = t^2 + \left(\frac{\pi^2}{4} - \pi t + t^2\right) = 2t^2 - \pi t + \frac{\pi^2}{4} \]

### Step 2: Finding the critical points

To find the maximum, we differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = 4t - \pi \]
Setting the derivative to zero:
\[ 4t - \pi = 0 \implies t = \frac{\pi}{4} \]

### Step 3: Evaluating \(y\) at the critical point

Substituting \(t = \frac{\pi}{4}\):
\[ y = 2\left(\frac{\pi}{4}\right)^2 - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4} \]
Calculating:
\[ y = 2 \cdot \frac{\pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8} \]
Thus, the maximum value \(a\) is:
\[ a = \frac{\pi^2}{8} \]

### Step 4: Finding the minimum value of \((\tan^{-1} x)^3 + (\cot^{-1} x)^3\)

Let:
\[ z = (\tan^{-1} x)^3 + (\cot^{-1} x)^3 \]
Using the identity:
\[ \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \]
We can express \(\cot^{-1} x\) in terms of \(\tan^{-1} x\):
\[ \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \]
Substituting this into \(z\):
\[ z = (\tan^{-1} x)^3 + \left(\frac{\pi}{2} - \tan^{-1} x\right)^3 \]
Let \(u = \tan^{-1} x\):
\[ z = u^3 + \left(\frac{\pi}{2} - u\right)^3 \]

### Step 5: Expanding and simplifying

Expanding the cube:
\[ z = u^3 + \left(\frac{\pi^3}{8} - \frac{3\pi^2}{4}u + 3\frac{\pi}{2}u^2 - u^3\right) \]
This simplifies to:
\[ z = \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 \]

### Step 6: Finding the critical points

Differentiating \(z\) with respect to \(u\):
\[ \frac{dz}{du} = -\frac{3\pi^2}{4} + 3\pi u \]
Setting the derivative to zero:
\[ -\frac{3\pi^2}{4} + 3\pi u = 0 \implies u = \frac{\pi}{4} \]

### Step 7: Evaluating \(z\) at the critical point

Substituting \(u = \frac{\pi}{4}\):
\[ z = \left(\frac{\pi}{4}\right)^3 + \left(\frac{\pi}{2} - \frac{\pi}{4}\right)^3 = \frac{\pi^3}{64} + \left(\frac{\pi}{4}\right)^3 = 2 \cdot \frac{\pi^3}{64} = \frac{\pi^3}{32} \]
Thus, the minimum value \(b\) is:
\[ b = \frac{\pi^3}{32} \]

### Step 8: Calculating \(a + \frac{b}{\pi}\)

Now we substitute \(a\) and \(b\):
\[ a + \frac{b}{\pi} = \frac{\pi^2}{8} + \frac{\frac{\pi^3}{32}}{\pi} = \frac{\pi^2}{8} + \frac{\pi^2}{32} \]
Finding a common denominator (32):
\[ = \frac{4\pi^2}{32} + \frac{\pi^2}{32} = \frac{5\pi^2}{32} \]

### Final Answer
Thus, the final result is:
\[ \frac{5\pi^2}{32} \]