Question

$\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$, then x =

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

Answer 1

Answer: (C)

$\frac{1}{6}$

Answer 2

To solve the equation $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$, we use the formula for the sum of arctangents:

$\tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$

In this case, $a = 2x$ and $b = 3x$. Substituting these values into the formula, we get:

$\tan^{-1}\left(\frac{2x+3x}{1-6x^2}\right) = \frac{\pi}{4}$

Since $\tan\left(\frac{\pi}{4}\right) = 1$, we have:

$\frac{5x}{1-6x^2} = 1$

Now, we solve the equation:

$5x = 1 - 6x^2$ $6x^2 + 5x - 1 = 0$

Using the quadratic formula:

$x = \frac{-5 \pm \sqrt{5^2 - 4\cdot 6\cdot (-1)}}{2\cdot 6} = \frac{-5 \pm \sqrt{25 + 24}}{12} = \frac{-5 \pm 7}{12}$

This gives us two possible solutions:

$x = \frac{2}{12} = \frac{1}{6}$ or $x = \frac{-12}{12} = -1$

We need to check the validity of these solutions.

For $x = \frac{1}{6}$, $(2x)(3x) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} < 1$, which satisfies the condition for the arctangent sum formula.

For $x = -1$, $(2(-1))(3(-1)) = 6 > 1$, and substituting in the original expression does not yield $\frac{\pi}{4}$.

Therefore, the valid solution is $x = \frac{1}{6}$.