Question

If $ta{n}^{-1}\frac{2x}{1-x^2}+co{t}^{-1}\frac{1-x^2}{2x}=\pi /3,x>0$ then $25(x^4+\frac{1}{x^4})+8$ is equal to

Answer 1

Simplifying, we get
$2ta{n}^{-1}\left(\frac{2x}{1-x^4}\right)=\frac{\pi }{3}$
$\Rightarrow ta{n}^{-1}\left(\frac{2x}{1-x^4}\right)=\frac{\pi }{6}$
$\Rightarrow \left(\frac{2x}{1-x^4}\right)=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{4x^2}{1+x^4-2x^2}=\frac{1}{3}$
$\Rightarrow 12x^2=1-2x^2+x^4$
$\Rightarrow 14x^2=x^4+1$
$\Rightarrow 196x^4=x^8+2x^4+1$
$\Rightarrow 194x^4=x^8+1$
$\Rightarrow 194=x^4+\frac{1}{x^4}$
Hence,
$25(x^4+\frac{1}{x^4})+8$
$=25\left(194\right)+8$
$=4850+8$
$=4858$