If  $\cos \left\{{\tan }^{-1}\left[\sin \left({\cot }^{-1}x\right)\right]\right\}=\sqrt{\frac{x^2+\alpha }{x^2+\beta }}$then$\frac{\alpha }{\beta }=$

$\begin{array}{l}We have {Cot}^{-1}x={\sin }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \mathrm{an}d {\tan }^{-1}x={\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\\ \therefore \cos \left\{{\tan }^{-1}\left[\sin \left(co{t}^{-1}x\right)\right]\right\}=\cos \left[{\tan }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right]\\ =\frac{1}{\sqrt{1+{\left(\frac{1}{\sqrt{1+x^2}}\right)}^2}}\\ =\sqrt{\frac{x^2+1}{x^2+2}}\\ =\sqrt{\frac{x^2+\alpha }{x^2+\beta }}\\ \therefore \frac{\alpha }{\beta }=\frac{1}{2}=0.5\end{array}$