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 and $250x^4+320x^2+137$ is equal to

Answer 1

Simplifying, we get
$ta{n}^{-1}\left(\frac{\frac{4}{x}}{1+x^2-\frac{4}{x^2}}\right)=ta{n}^{-1}\left(\frac{4}{x}\right)$
$\Rightarrow 1+x^2-\frac{4}{x^2}=1$
$\Rightarrow x^2-\frac{4}{x^2}=0$
$\Rightarrow x^2=\frac{4}{x^2}$
$\Rightarrow x^4=4$
$\Rightarrow x^2=2$ ... for real roots
Hence
$250x^4+320x^2+137$
$=250\left(4\right)+320\left(2\right)+137$
$=1000+640+137$
$=1777$