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: Read the equation.
We solve $\tan^{-1}\!\left(\frac{x-1}{x-2}\right)+\tan^{-1}\!\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$.
Step 2: Use the addition rule.
Combine the two arctangents with $\tan^{-1}A+\tan^{-1}B=\tan^{-1}\!\left(\frac{A+B}{1-AB}\right)$. Then take tangent of both sides, so $\frac{A+B}{1-AB}=\tan\frac{\pi}{4}=1$.
Step 3: Add the numerators.
$A+B=\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}=\frac{(x^2+x-2)+(x^2-x-2)}{(x-2)(x+2)}=\frac{2x^2-4}{(x-2)(x+2)}$.
Step 4: Work out the denominator 1-AB.
$1-AB=\frac{(x-2)(x+2)-(x-1)(x+1)}{(x-2)(x+2)}=\frac{(x^2-4)-(x^2-1)}{(x-2)(x+2)}=\frac{-3}{(x-2)(x+2)}$.
Step 5: Form the equation.
Dividing, the common denominator cancels: $\frac{2x^2-4}{-3}=1$, so $2x^2-4=-3$.
Step 6: Solve for x.
Then $2x^2=1$, so $x^2=\tfrac{1}{2}$ and $x=\pm\tfrac{1}{\sqrt{2}}$, which is option (C).
\[ \boxed{\,x=\pm\tfrac{1}{\sqrt{2}}\,} \]