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: Understanding the Question:
The question requires us to find the positive value of $x$ that satisfies the given inverse trigonometric equation.

Step 2: Key Formula or Approach:
We will use the standard inverse trigonometric identity for adding arc-tangent functions:
$$\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right)$$

Step 3: Detailed Explanation:
Apply the identity setting $A = 2x$ and $B = 3x$:
$$\tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) = \frac{\pi}{4}$$ $$\frac{5x}{1 - 6x^2} = \tan\left(\frac{\pi}{4}\right)$$ Since $\tan\left(\frac{\pi}{4}\right) = 1$:
$$\frac{5x}{1 - 6x^2} = 1 \implies 5x = 1 - 6x^2$$ Rearrange this into a standard quadratic equation form:
$$6x^2 + 5x - 1 = 0$$ Factor the quadratic equation by splitting the middle term:
$$6x^2 + 6x - x - 1 = 0$$ $$6x(x + 1) - 1(x + 1) = 0 \implies (6x - 1)(x + 1) = 0$$ This yields two potential mathematical roots: $x = \frac{1}{6}$ or $x = -1$.
Since the problem explicitly restricts the domain to $x > 0$, we discard the negative root. Thus, $x = \frac{1}{6}$, matching option (B).

Step 4: Final Answer:
The value of $x$ is $\frac{1}{6}$, which corresponds to option (B).