# Tangent is one-to-one with a restricted domain.

I want to show the function $\tan(\pi x - \frac \pi2)$ is one-to-one if x $\in (0,1)$. But an argument I would normally use to prove a function is one-to-one (letting $\tan(\pi x_1 - \frac \pi2)$ = $\tan(\pi x_2 - \frac \pi2)$ and then showing $x_1$ = $x_2$) doesn't seem to work.

I was hoping someone could give me any suggestions.

Thanks.

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## 2 Answers

The derivative is defined (exists) and positive on $(0.1)$, so the function is increasing over the interval.

Remark: The argument does not work over larger intervals, since then the derivative does not always exist, although it is always positive where it exists.

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Thank you very much. – mkeachie Oct 2 '12 at 5:02

If $x\in(0,1)$, then $0<x<1$, so $-\frac{\pi}2<\pi x-\frac{\pi}2<\frac{\pi}2$. Note that $\tan\theta$ is one-to-one on the interval $-\frac{\pi}2<\theta<\frac{\pi}2$: each $\theta$ determines a unique line through the origin making an angle $\theta$ with the positive $x$-axis, the point $\langle 1,\tan\theta\rangle$ is on that line, and these lines meet only at the origin, so these points are all distinct, and therefore the values of $\tan\theta$ are all distinct.

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