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: Combine the two arctan terms.
Use $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\tfrac{A+B}{1-AB}$ with $A=\tfrac{x-1}{x-2}$ and $B=\tfrac{x+1}{x+2}$. Since the right side equals $\tfrac{\pi}{4}$, the fraction inside must equal $1$.

Step 2: Build the numerator.
$A+B$ over a common base has top $(x-1)(x+2)+(x+1)(x-2) = (x^2+x-2)+(x^2-x-2) = 2x^2-4$.

Step 3: Build the denominator.
$1-AB$ over the same base has top $(x-2)(x+2)-(x-1)(x+1) = (x^2-4)-(x^2-1) = -3$. So the equation is $\tfrac{2x^2-4}{-3} = 1$.

Step 4: Solve for $x$.
Then $2x^2-4 = -3$, so $2x^2 = 1$ and $x^2 = \tfrac12$. \[ \boxed{x = \pm\tfrac{1}{\sqrt2}} \]