Question

If $\tan^{-1}\left(\frac{2}{3 - x + 1}\right) = \cot^{-1}\left(\frac{3}{3x + 1}\right)$, then which one of the following is true?

• A. There is no real value of $x$ satisfying the above equation.
• B. There is one positive and one negative real value of $x$ satisfying the above equation.
• C. There are two real positive values of $x$ satisfying the above equation.
• D. There are two real negative values of $x$ satisfying the above equation.

Answer 1

Answer: (B)

There is one positive and one negative real value of $x$ satisfying the above equation.

Answer 2

Given the equation:

$\tan^{-1} \left( \frac{2}{3x + 1} \right) = \cot^{-1} \left( \frac{3}{3x + 1} \right).$

We know that $\cot^{-1} y = \frac{\pi}{2} - \tan^{-1} y$. Thus, we can rewrite the equation as:

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

Now, use the identity $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right)$ for $a = \frac{2}{3x + 1}$ and $b = \frac{3}{3x + 1}$:

$\tan^{-1} \left( \frac{2}{3x + 1} \right) + \tan^{-1} \left( \frac{3}{3x + 1} \right) = \tan^{-1} \left( \frac{\frac{2}{3x+1} + \frac{3}{3x+1}}{1 - \frac{2}{3x+1} \cdot \frac{3}{3x+1}} \right).$

Simplify this expression to find the values of $x$, resulting in two solutions: one positive and one negative.

Thus, the correct answer is: (2) There is one positive and one negative real value of $x$.