Question

If \((tan^{-1}x)^2+(cot^{-1}x)^2 = \frac{5π^2}{8}\), then the value of \(x\) is equal to...

• A. \(-1\)
• B. \(-2\)
• C. \(1\)
• D. \(2\)

Hint:

Use cot^-1 x = pi/2 - tan^-1 x and solve the quadratic in tan^-1 x.

Answer 1

The Correct Option is A

Step 1: Substitute
Let \(t = \tan^{-1}x\), so \(t\in(-\frac\pi2,\frac\pi2)\). Also \(\cot^{-1}x = \frac\pi2 - t\).

Step 2: Form the quadratic
\(t^2+\left(\frac\pi2-t\right)^2 = \frac{5\pi^2}{8}\) gives \(2t^2 - \pi t + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0\), that is \(16t^2 - 8\pi t - 3\pi^2 = 0\).

Step 3: Solve
\(t = \frac{8\pi\pm\sqrt{64\pi^2+192\pi^2}}{32} = \frac{8\pi\pm16\pi}{32}\), so \(t = \frac{3\pi}{4}\) or \(t = -\frac\pi4\).

Step 4: Select
\(t = \frac{3\pi}{4}\) is outside the range of \(\tan^{-1}\), so \(t = -\frac\pi4\) and \(x = \tan(-\frac\pi4) = -1\). Option (A).

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