The value of ${\left(\frac{1}{y^2}{\left(\frac{\cos \left({\tan }^{-1}y\right)+y\sin \left({\tan }^{-1}y\right)}{\cot \left({\sin }^{-1}y\right)+\tan \left({\sin }^{-1}y\right)}\right)}^2+y^4\right)}^{1/2}$ is……..

We have 

$\left(\frac{\cos \left({\tan }^{-1}y\right)+y\sin \left({\tan }^{-1}y\right)}{\cot \left({\sin }^{-1}y\right)+\tan \left({\sin }^{-1}y\right)}\right)$

$=\left(\frac{\cos \left({\cos }^{-1}\left(\frac{1}{\sqrt{y^2+1}}\right)\right)+y\sin \left({\sin }^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)\right)}{\cot \left({\cot }^{-1}\left(\frac{\sqrt{1-y^2}}{y}\right)\right)+\tan \left({\tan }^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right)\right)}\right)$

$=\left(\frac{\left(\frac{1}{\sqrt{y^2-1}}\right)+y\left(\frac{y}{1+y^2}\right)}{\left(\frac{\sqrt{1-y^2}}{y}\right)+\left(\frac{y}{\sqrt{1-y^2}}\right)}\right)$

$=\left(\frac{\left(\frac{1+y^2}{\sqrt{y^2+1}}\right)}{\left(\frac{1-y^2+y^2}{\sqrt{y-y^2}}\right)}\right)$

$=y\left(\sqrt{1-y^4}\right)$

Thus, the given expression reduces to 

$={(\frac{1}{y^2}{\left(y\sqrt{1-y^4}\right)}^2+y^4)}^{\frac{1}{2}}$

$={(\frac{y^4(1-y^4)}{y^2}+y^4)}^{\frac{1}{2}}$

$={((1-y^4)+y^4)}^{1/2}$

$=1$