Simple inverse trigo....

$\sin(\cot^{-1}(\tan(\cos^{-1}x))) \\ Let, \cos^{-1}x = \theta \\ So, \cos \theta = x \\ Then, \sin \theta = \displaystyle \sqrt{(1 - x^2)} \\ Then, \tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{\sqrt{1 - x^2}}{x} \\ So, {\theta = \tan^{-1} \dfrac{\sqrt{1 - x^2}}{x}} \\ \therefore {\color{#3D99F6}{cos^{-1}}x = \tan^{-1} \dfrac{\sqrt{1 - x^2}}{x} = \theta}. \\ {\text{Putting this in the question , we get -}} \\ A = \sin(\cot^{-1}(\tan(\cos^{-1}x))) = \sin(\cot^{-1}(\tan(\tan^{-1} \dfrac{\sqrt{1 - x^2}}{x}))) = \sin(\cot^{-1}(\dfrac{\sqrt{1 - x^2}}{x}))............................(1) \\ Let, \cot^{-1}(\dfrac{\sqrt{1 - x^2}}{x}) = \phi \\ So, \cot \phi = \dfrac{\sqrt{1 - x^2}}{x} \\ \dfrac{\cos \phi}{\sin \phi} = \dfrac{\sqrt{1 - x^2}}{x} \\ \implies \dfrac{\cos^{2} \phi}{\sin^{2} \phi} = \dfrac{1 - x^2}{x^2} \implies \dfrac{1 - \sin^2 \phi}{\sin^2 \phi} = \dfrac{1 - x^2}{x^2} \implies \sin \phi = x \implies \sin^{-1} x = \phi \\ \color{#3D99F6}{\therefore \sin^{-1} x} = \cot^{-1}(\dfrac{\sqrt{1 - x^2}}{x}) =\phi \\ {\text{Putting this in 1, we get -}} \\ A = \sin(\cot^{-1}(\dfrac{\sqrt{1 - x^2}}{x})) = \sin(\sin x) = \huge \boxed{x}$