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(D) Two functions are identical if they have the same domain and the same rule of correspondence.$(A)$ For $y = 	an(\cos^{-1} x)$,the domain of $\cos

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(D) Two functions are identical if they have the same domain and the same rule of correspondence.$(A)$ For $y = 	an(\cos^{-1} x)$,the domain of $\cos^{-1} x$ is $[-1, 1]$. Since $	an(	heta)$ is undefined at $	heta = \frac{\pi}{2}$,we must have $\cos^{-1} x 
eq \frac{\pi}{2}$,which implies $x 
eq 0$. Thus,the domain is $[-1, 0) \cup (0, 1]$. The expression simplifies to $\frac{\sqrt{1-x^2}}{x}$. Both functions share the same domain and expression.$(B)$ For $y = 	an(\cot^{-1} x)$,since $\cot^{-1} x$ is defined for all $x \in \mathbb{R}$ and $\cot^{-1} x \in (0, \pi)$,$	an(\cot^{-1} x)$ is defined for all $x \in \mathbb{R}$ except where $\cot^{-1} x = \frac{\pi}{2}$,i.e.,$x = 0$. Thus,the domain is $\mathbb{R} \setminus \{0\}$. The expression simplifies to $\frac{1}{x}$. Both functions are identical.$(C)$ For $y = \sin(	an^{-1} x)$,the domain of $	an^{-1} x$ is $\mathbb{R}$. Since $	an^{-1} x \in (-\frac{\pi}{2}, \frac{\pi}{2})$,$\sin(	an^{-1} x)$ is defined for all $x \in \mathbb{R}$. The expression simplifies to $\frac{x}{\sqrt{1+x^2}}$. Both functions are identical.Since $(A)$,$(B)$,and $(C)$ are all pairs of identical functions,the correct option is $(D)$.

Identify the pair$(s)$ of functions which are identical.

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