Identify the pair(s) of functions which are identical

$\text{ a) }\mathrm{y}=\tan \left({\cos }^{-1}\mathrm{x}\right)\Rightarrow 0\le {\cos }^{-1}\mathrm{x}\le \mathrm{\pi }$
$$\begin{gathered} \mathrm{\&}{\cos }^{-1}\mathrm{x}\ne \frac{\mathrm{\pi }}{2}\Rightarrow -1\le \mathrm{x}\le 1\mathrm{\&}\mathrm{x}\ne 0 \\ y=\tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)=\frac{\sqrt{1-x^2}}{x} \end{gathered}$$

$\text{ b) }\mathrm{y}=\tan \left({\cot }^{-1}\mathrm{x}\right)\Rightarrow 0<{\cot }^{-1}\mathrm{x}<\mathrm{\pi }$
$$\begin{gathered} \mathrm{\&}{\cos }^{-1}\mathrm{x}\ne \frac{\mathrm{\pi }}{2}\therefore \mathrm{x}\in \mathrm{R}-\left\{0\right\} \\ \mathrm{y}=\tan \left({\cot }^{-1}\mathrm{x}\right)=\tan \left({\tan }^{-1}\frac{1}{x}\right)=\frac{1}{x} \end{gathered}$$
$\text{ c) }\mathrm{y}=\sin \left({\tan }^{-1}\mathrm{x}\right)\Rightarrow -\frac{\mathrm{\pi }}{2}<{\tan }^{-1}\mathrm{x}<\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{x}\in \mathrm{R}\text{; }$
$$\begin{gathered} \therefore \mathrm{y}\in (-1,1) \\ \mathrm{y}=\sin \left({\tan }^{-1}\mathrm{x}\right)=\sin \left({\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)=\frac{x}{\sqrt{1+x^2}} \end{gathered}$$

$\text{ d) }\mathrm{y}=\cos \left({\tan }^{-1}\mathrm{x}\right)\Rightarrow \mathrm{y}\in (0,1]\mathrm{\&}$
$\mathrm{x}\in \mathrm{R}\Rightarrow \mathrm{y}=\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{\&}\mathrm{y}=\sin \left({\cot }^{-1}\mathrm{x}\right)$ has same graph