For all real numbers x except x=0 and x=1 the function f(x) is defined by f(xx−1)=1x Suppose 0≤t≤π2. What is the value of f(sec2t)? I first set sec2t=xx−1 and then plugged it in to WA which told me that x=csc2t so that would mean that 1x=1csc2x. However, that turned out to be wrong, so what should I do here?

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For all real numbers x except x=0 and x=1 the function f(x) is defined by  f(xx−1)=1x  Suppose 0≤t≤π2. What is the value of f(sec2t)?   I first set sec2t=xx−1 and then plugged it in to WA which told me that x=csc2t so that would mean that 1x=1csc2x. However, that turned out to be wrong, so what should I do here?

Cassandra Ramirez · Accepted Answer

Hint: let g(x)=xx−1. Then f(g(x))=1x, so f(x)=1g−1(x).  That inverse function is not hard to compute.

RizerMix · Accepted Answer

We have f(1cos2⁡θ)=f(11−sin2⁡θ)=f(csc2⁡θcsc2⁡θ−1)=1csc2⁡θ=sin2⁡θ

user_27qwe · Accepted Answer

You mean 1csc2⁡t=sin2⁡t as (change t to θ) per first answer Alternatively, since f(y)=1−1y f(sec2⁡t)=1−cos2⁡t=sin2⁡t

For all real numbers x except x=0 and x=1 the function f(x) is defined by f(\frac{x}{x-1})=\f

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