Question

$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$, then the values of $x$ are

• A. $\pm \frac{3}{\sqrt{2}}$
• B. $\pm \frac{1}{2}$
• C. $\pm \frac{1}{\sqrt{2}}$
• D. $\pm \frac{\sqrt{3}}{2}$

Hint:

Always check the domain of the arctangent functions when solving, though here the algebraic simplification confirms the valid roots.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given an inverse trigonometric equation involving a sum of arctangent functions equal to $\frac{\pi}{4}$. We need to solve for $x$.

Step 2: Key Formula or Approach:
Use the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$.

Step 3: Detailed Explanation:
Let $A = \frac{x-1}{x-2}$ and $B = \frac{x+1}{x+2}$.
$\tan^{-1} \left( \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - (\frac{x-1}{x-2})(\frac{x+1}{x+2})} \right) = \frac{\pi}{4}$
$\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = \tan\left(\frac{\pi}{4}\right) = 1$
$\frac{(x^2+x-2) + (x^2-x-2)}{(x^2-4) - (x^2-1)} = 1$
$\frac{2x^2 - 4}{-3} = 1 \implies 2x^2 - 4 = -3 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2}$
Thus, $x = \pm \frac{1}{\sqrt{2}}$.

Step 4: Final Answer:
The values of $x$ are $\pm \frac{1}{\sqrt{2}}$, which corresponds to option (C).