Question

If $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = 5\pi^2/8$, then $x^2 + 1 = \dots$

• A. -1
• B. 2
• C. 1
• D. -2

Hint:

Always check if your calculated inverse trigonometric angles fall within their universally defined principal domains! Failing to reject $3\pi/4$ here could lead to confusion, even though $\tan(3\pi/4)$ also coincidentally evaluates to -1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use the identity $\tan^{-1} x + \cot^{-1} x = \pi/2$. Let $\tan^{-1} x = \theta$. Then $\cot^{-1} x = \pi/2 - \theta$.

Step 2: Formula Application:
The given equation becomes: $\theta^2 + (\pi/2 - \theta)^2 = 5\pi^2/8$. $\theta^2 + \pi^2/4 + \theta^2 - \pi\theta = 5\pi^2/8$.

Step 3: Explanation:
$2\theta^2 - \pi\theta + \pi^2/4 - 5\pi^2/8 = 0 \implies 2\theta^2 - \pi\theta - 3\pi^2/8 = 0$. Multiplying by 8: $16\theta^2 - 8\pi\theta - 3\pi^2 = 0$. Factoring: $(4\theta - 3\pi)(4\theta + \pi) = 0$. $\theta = 3\pi/4$ or $\theta = -\pi/4$. If $\tan^{-1} x = -\pi/4$, then $x = -1$. Then $x^2 + 1 = (-1)^2 + 1 = 2$.

Step 4: Final Answer:
The value of $x^2 + 1$ is 2.