To solve the given equation: \(\text{Tan}^{-1}(x+\frac{\sqrt{2}}{x}) + \text{Tan}^{-1}(x-\frac{\sqrt{2}}{x}) = \text{Tan}^{-1}(x)\), we need to simplify and analyze the given trigonometric relationships.
First, we use the formula for the addition of inverse tangents: \(\text{Tan}^{-1} a + \text{Tan}^{-1} b = \text{Tan}^{-1} \left(\frac{a + b}{1 - ab}\right)\) if \(ab < 1\).
Let \(a = x + \frac{\sqrt{2}}{x}\) and \(b = x - \frac{\sqrt{2}}{x}\).
Calculate: \(a + b = \left(x + \frac{\sqrt{2}}{x}\right) + \left(x - \frac{\sqrt{2}}{x}\right) = 2x\)
and \(ab = \left(x + \frac{\sqrt{2}}{x}\right)\left(x - \frac{\sqrt{2}}{x}\right)\) which simplifies to:
\(ab = x^2 - \left(\frac{\sqrt{2}}{x}\right)^2 = x^2 - \frac{2}{x^2}\)
Now, substituting these into the formula: \(\text{Tan}^{-1}(2x) = \text{Tan}^{-1}\left(\frac{2x}{1 - \left(x^2 - \frac{2}{x^2}\right)}\right)\)
Thus, the equation becomes: \(\text{Tan}^{-1}(x)\) = \(\text{Tan}^{-1}\left(\frac{2x}{\left(\frac{x^4 - x^2 + 2}{x^2}\right)}\right)\)
Equating the arguments of the inverse tangents, we have: \(x = \frac{2x \cdot x^2}{x^4 - x^2 + 2}\)
This simplifies to: \(x^4 - x^3 - x^2 + 2 = 0\)
Checking for possible roots, notice that for certain values of \(x\), the equation simplifies to standard roots. By testing or factorization methods:
The possible solutions to the polynomial can be determined as \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). Both satisfy the original equation yielding two valid solutions.
Therefore, the number of values of \(x\) that satisfy the given equation is 2.