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 Question:
We are given an equation involving the squares of inverse tangent and inverse cotangent. We must find the value of $x$, and then evaluate $x^2 + 1$.

Step 2: Key Formula or Approach:
Use the fundamental inverse trigonometric identity:
$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all real $x$.
We can substitute $\cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x$ into the given equation to form a quadratic equation in terms of a single variable, $\tan^{-1} x$.

Step 3: Detailed Explanation:
Let $A = \tan^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - A$.
Substitute these into the given equation:
$$A^2 + \left(\frac{\pi}{2} - A\right)^2 = \frac{5\pi^2}{8}$$
Expand the squared binomial:
$$A^2 + \frac{\pi^2}{4} - \pi A + A^2 = \frac{5\pi^2}{8}$$
Combine like terms:
$$2A^2 - \pi A + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0$$
Find a common denominator for the constants (which is 8):
$$2A^2 - \pi A + \frac{2\pi^2 - 5\pi^2}{8} = 0$$
$$2A^2 - \pi A - \frac{3\pi^2}{8} = 0$$
Multiply the entire equation by 8 to clear the fraction:
$$16A^2 - 8\pi A - 3\pi^2 = 0$$
Factor the quadratic equation by splitting the middle term. We need two numbers that multiply to $16 \times -3\pi^2 = -48\pi^2$ and add to $-8\pi$. Those numbers are $-12\pi$ and $+4\pi$.
$$16A^2 - 12\pi A + 4\pi A - 3\pi^2 = 0$$
$$4A(4A - 3\pi) + \pi(4A - 3\pi) = 0$$
$$(4A + \pi)(4A - 3\pi) = 0$$
This yields two possible solutions for $A$:
$A = -\pi/4$ or $A = 3\pi/4$
Recall that $A = \tan^{-1} x$. The principal value branch of $\tan^{-1} x$ is strictly strictly within the interval $(-\pi/2, \pi/2)$.
Therefore, $A = 3\pi/4$ is an extraneous root and must be rejected.
We are left with:
$$\tan^{-1} x = -\frac{\pi}{4}$$
Taking the tangent of both sides:
$$x = \tan\left(-\frac{\pi}{4}\right) = -1$$
Finally, evaluate the required expression $x^2 + 1$:
$$(-1)^2 + 1 = 1 + 1 = 2$$

Step 4: Final Answer:
The value of $x^2 + 1$ is 2, matching option (b).