Which of the following pairs of functions are | vedclass

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(D) Let us analyze each pair:$1$. For option $A$: $g(x) = \cot^2 x - \cos^2 x = \frac{\cos^2 x}{\sin^2 x} - \cos^2 x = \cos^2 x \left( \frac{1}{\sin^2

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All of the above

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(D) Let us analyze each pair:$1$. For option $A$: $g(x) = \cot^2 x - \cos^2 x = \frac{\cos^2 x}{\sin^2 x} - \cos^2 x = \cos^2 x \left( \frac{1}{\sin^2 x} - 1 ight) = \cos^2 x \left( \frac{1 - \sin^2 x}{\sin^2 x} ight) = \cos^2 x \cdot \frac{\cos^2 x}{\sin^2 x} = \cos^2 x \cdot \cot^2 x = f(x)$. Since the domains are also identical $(x 
eq n\pi)$,these functions are identical.$2$. For option $B$: $f(x) = 	an(	an^{-1} x) = x$ for all $x \in \mathbb{R}$. $g(x) = \cot(\cot^{-1} x) = x$ for all $x \in \mathbb{R}$. Since both functions are $y = x$ for all $x \in \mathbb{R}$,they are identical.$3$. For option $C$: $f(x) = 	ext{sgn}(x)$. $g(x) = 	ext{sgn}(	ext{sgn}(x))$. If $x > 0$,$f(x) = 1$ and $g(x) = 	ext{sgn}(1) = 1$. If $x = 0$,$f(x) = 0$ and $g(x) = 	ext{sgn}(0) = 0$. If $x Since all pairs are identical,the correct option is $D$.

Which of the following pairs of functions are identical?

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