Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given $x\times y=x^2+y^3$

Now $\left(x\times 1\right)\times 1=x\times \left(1\times 1\right)$ 

$\left(x^2+1^3\right)\times 1=x\times \left(1^2+1^3\right)$ 

${\left(x^2+1\right)}^2+1^3=x^2+2^3$

$x^4+1+2x^2+1=x^2+8$ 

$x^4+x^2-6=0$

Let $x^2=t$   

$t^2+t-6=0$ 

$\left(t+3\right)\left(t-2\right)=0$

$t\ne -3$ so $t=2$ or $x^2=2$

Now $2{\sin }^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)=2{\sin }^{-1}\left(\frac{4+2-2}{4+2+2}\right)$
$=2{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}\times 2=\frac{\mathrm{\pi }}{3}$