Question

If $ta{n}^{-1}(x+\frac{2}{x})-ta{n}^{-1}(x-\frac{2}{x})=ta{n}^{-1}\frac{4}{x}$ then the value of $2x^4-3x^2$ is equal to

• A. $2$
• B. $0$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A $2$

$ta{n}^{-1}(x+\frac{2}{x})-ta{n}^{-1}(x-\frac{2}{x})$
$=ta{n}^{-1}\left(\frac{x+\frac{2}{x}-x+\frac{2}{x}}{1+(x+\frac{2}{x}(x-\frac{2}{x}\left)\right)}\right)$
$=ta{n}^{-1}\left(\frac{\frac{4}{x}}{1+x^2-\frac{4}{x^2}}\right)$
$=ta{n}^{-1}\left(\frac{4}{x}\right)$
Hence
$1+x^2-\frac{4}{x^2}=1$
$x^2-\frac{4}{x^2}=0$
$x^4=4$
$x^2=2$ for all real x.
Hence $2x^4-3x^2$
$=2\left(4\right)-3\left(2\right)$
$=8-6$
$=2$