If `(tan^(-1)x)^(2)+(cot^(-1)x)^(2)=(5 pi^(2))/(8)`, then the value of x is

Options

A. `-2`
B. `-3`
C. `-1`
D. 2

Answer

To solve the equation \((\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}\), we can follow these steps:

### Step 1: Use the identity for \(\cot^{-1} x\)
We know that:
\[ \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \]
Substituting this into the equation gives:
\[ (\tan^{-1} x)^2 + \left(\frac{\pi}{2} - \tan^{-1} x\right)^2 = \frac{5\pi^2}{8} \]

### Step 2: Expand the equation
Now, let's expand the second term:
\[ \left(\frac{\pi}{2} - \tan^{-1} x\right)^2 = \left(\frac{\pi}{2}\right)^2 - 2\left(\frac{\pi}{2}\right)(\tan^{-1} x) + (\tan^{-1} x)^2 \]
This simplifies to:
\[ \frac{\pi^2}{4} - \pi \tan^{-1} x + (\tan^{-1} x)^2 \]
Now substituting this back into the equation:
\[ (\tan^{-1} x)^2 + \frac{\pi^2}{4} - \pi \tan^{-1} x + (\tan^{-1} x)^2 = \frac{5\pi^2}{8} \]
Combining like terms gives:
\[ 2(\tan^{-1} x)^2 - \pi \tan^{-1} x + \frac{\pi^2}{4} = \frac{5\pi^2}{8} \]

### Step 3: Rearranging the equation
Now, let's move \(\frac{5\pi^2}{8}\) to the left side:
\[ 2(\tan^{-1} x)^2 - \pi \tan^{-1} x + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0 \]
Finding a common denominator (which is 8):
\[ \frac{2\pi^2}{8} - \frac{5\pi^2}{8} = -\frac{3\pi^2}{8} \]
Thus, we have:
\[ 2(\tan^{-1} x)^2 - \pi \tan^{-1} x - \frac{3\pi^2}{8} = 0 \]

### Step 4: Solve the quadratic equation
Let \(y = \tan^{-1} x\). The equation becomes:
\[ 2y^2 - \pi y - \frac{3\pi^2}{8} = 0 \]
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- \(a = 2\)
- \(b = -\pi\)
- \(c = -\frac{3\pi^2}{8}\)

Calculating the discriminant:
\[ b^2 - 4ac = \pi^2 - 4 \cdot 2 \cdot \left(-\frac{3\pi^2}{8}\right) = \pi^2 + 3\pi^2 = 4\pi^2 \]
Now substituting back into the quadratic formula:
\[ y = \frac{\pi \pm 2\pi}{4} \]
This gives us two solutions:
1. \(y = \frac{3\pi}{4}\)
2. \(y = -\frac{\pi}{4}\)

### Step 5: Determine the valid solution
Since \(y = \tan^{-1} x\) must lie between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), we discard \(y = \frac{3\pi}{4}\) because it is outside this range. Thus:
\[ y = -\frac{\pi}{4} \]
This implies:
\[ \tan^{-1} x = -\frac{\pi}{4} \implies x = \tan\left(-\frac{\pi}{4}\right) = -1 \]

### Final Answer
The value of \(x\) is:
\[ \boxed{-1} \]