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Let f: R->R be given by f(x)=tanx . Then...

Let `f: R->R` be given by `f(x)=tanx` . Then, `f^(-1)(1)` is `pi/4` (b) `{npi+pi/4: n in Z}` (c) does not exist (d) none of these

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Given,
`f: R->R` be given by `f(x)=tanx`
Since `f ^-1` is inverse of` f` .
`=>(f0f^-1)(1)=1`
`=>f(f^-1(1))=1`
`=>tan f(f^-1(1))=1`
but, `tan(pi/4) =1`
`=>tan f(f^-1(1))=tan(pi/4)`
`=>f^-1=npi+pi/4(ninz)`
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