Question

The number of solutions of $\tan^{-1} (x + \frac{2}{x}) - \tan^{-1} (\frac{4}{x}) - \tan^{-1} (x - \frac{2}{x}) = 0$ are

• A. 1
• B. 2
• C. 3
• D. 0

Hint:

$\tan^{-1} A - \tan^{-1} B = \tan^{-1} \frac{A-B}{1+AB}$.

Answer 1

The Correct Option is B

Step 1: Rearrange Equation
$\tan^{-1} (x + \frac{2}{x}) - \tan^{-1} (x - \frac{2}{x}) = \tan^{-1} (\frac{4}{x})$.
Step 2: Apply Formula
$\tan^{-1} \left[ \frac{(x + \frac{2}{x}) - (x - \frac{2}{x})}{1 + (x + \frac{2}{x})(x - \frac{2}{x})} \right] = \tan^{-1} (\frac{4}{x})$.
$\frac{4/x}{1 + x^2 - 4/x^2} = \frac{4}{x}$.
Step 3: Solve for x
Assuming $x \neq 0$, we have $1 = 1 + x^2 - \frac{4}{x^2} \implies x^2 = \frac{4}{x^2} \implies x^4 = 4$.
$x^2 = 2 \implies x = \pm \sqrt{2}$.
There are 2 real solutions.
Final Answer: (B)