Question

If $\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}$, where $x > 0$, then $x =$

• A. $1$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Hint:

When doing multiple-choice math, substituting options can save time! If you quickly plug in $x = \frac{1}{6}$, the equation becomes $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right)$. Using the formula gives $\tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1) = \frac{\pi}{4}$, instantly verifying the answer!

Answer 1

The Correct Option is B

Step 1: Recall the addition rule for arctangents.
$\tan^{-1}A + \tan^{-1}B = \tan^{-1}\dfrac{A+B}{1-AB}$ when $AB < 1$.
Step 2: Apply it with $A = 2x$ and $B = 3x$.
$\tan^{-1}\dfrac{2x+3x}{1-(2x)(3x)} = \dfrac{\pi}{4}$, that is $\tan^{-1}\dfrac{5x}{1-6x^2} = \dfrac{\pi}{4}$.
Step 3: Take the tangent of both sides.
Since $\tan\dfrac{\pi}{4} = 1$, we get $\dfrac{5x}{1-6x^2} = 1$.
Step 4: Clear the fraction.
$5x = 1 - 6x^2$, which rearranges to $6x^2 + 5x - 1 = 0$.
Step 5: Factor the quadratic.
Splitting the middle term: $6x^2 + 6x - x - 1 = 0 \Rightarrow (6x - 1)(x + 1) = 0$. So $x = \dfrac{1}{6}$ or $x = -1$.
Step 6: Apply the given restriction.
Since $x > 0$, we reject $x = -1$ and keep $x = \dfrac{1}{6}$.
\[ \boxed{x = \dfrac{1}{6}} \]