Question

The number of values of x satisfying the equation $\text{Tan}^{-1}(x+\frac{\sqrt{2}}{x}) + \text{Tan}^{-1}(x-\frac{\sqrt{2}}{x}) = \text{Tan}^{-1}(x)$ is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When using the addition formula for $\text{Tan}^{-1}$, always state the condition under which it's valid (e.g., $AB<1$) and check your final solutions against this condition.

Answer 1

The Correct Option is C

The given equation is $\text{Tan}^{-1}(A) + \text{Tan}^{-1}(B) = \text{Tan}^{-1}(C)$, where $A=x+\frac{\sqrt{2}}{x}$, $B=x-\frac{\sqrt{2}}{x}$, and $C=x$.
We use the formula $\text{Tan}^{-1}(A) + \text{Tan}^{-1}(B) = \text{Tan}^{-1}\left(\frac{A+B}{1-AB}\right)$, which is valid if $AB<1$. 
Let's check the product $AB$: $AB = (x+\frac{\sqrt{2}}{x})(x-\frac{\sqrt{2}}{x}) = x^2 - (\frac{\sqrt{2}}{x})^2 = x^2 - \frac{2}{x^2}$. 
Applying the formula to the left side of the equation: 
$\text{Tan}^{-1}\left(\frac{(x+\frac{\sqrt{2}}{x}) + (x-\frac{\sqrt{2}}{x})}{1-(x^2 - \frac{2}{x^2})}\right) = \text{Tan}^{-1}(x)$. 
Taking tan of both sides: 
$\frac{2x}{1 - x^2 + \frac{2}{x^2}} = x$. 
Since $x=0$ is not in the domain of the original equation, we can assume $x\neq0$ and divide both sides by $x$. 
$\frac{2}{1 - x^2 + \frac{2}{x^2}} = 1$. 
$2 = 1 - x^2 + \frac{2}{x^2}$. 
$1 = -x^2 + \frac{2}{x^2}$. 
Multiply by $x^2$ and rearrange into a quadratic form in $x^2$: 
$x^2 = -x^4 + 2 \implies x^4 + x^2 - 2 = 0$. 
Let $y=x^2$. The equation becomes $y^2+y-2=0$. 
Factoring gives $(y+2)(y-1)=0$. 
So, $y=-2$ or $y=1$. 
Since $y=x^2$, $y$ cannot be negative. Thus, we only consider $y=1$. 
$x^2 = 1 \implies x = 1$ or $x = -1$. 
We must verify the condition $AB<1$ for these solutions. For $x=1$ or $x=-1$, we have $x^2=1$. $AB = x^2 - \frac{2}{x^2} = 1 - \frac{2}{1} = -1$. Since $-1<1$, the formula we used was valid for these solutions. 
There are two values of x that satisfy the equation.