Let f: R rightarrow R be given by f(x) = tan | vedclass

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(B) Given $f(x) = 	an x$.The inverse function $f^{-1}(y)$ is defined such that $f(f^{-1}(y)) = y$.Here,we need to find $f^{-1}(1)$,so we set $f(x) =

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$\{n \pi + \frac{\pi}{4} : n \in Z\}$

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(B) Given $f(x) = 	an x$.The inverse function $f^{-1}(y)$ is defined such that $f(f^{-1}(y)) = y$.Here,we need to find $f^{-1}(1)$,so we set $f(x) = 1$.$	an x = 1$.The general solution for $	an x = 	an \alpha$ is $x = n \pi + \alpha$,where $n \in Z$.Since $	an(\frac{\pi}{4}) = 1$,the set of values for $x$ is $\{n \pi + \frac{\pi}{4} : n \in Z\}$.Therefore,$f^{-1}(1) = \{n \pi + \frac{\pi}{4} : n \in Z\}$.

Let $f: R \rightarrow R$ be given by $f(x) = \tan x$. Then,$f^{-1}(1)$ is

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