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I've the following parametric equations for a curve: $$\begin{cases}x(t)=a(t-\tanh(t))\\y(t)=a \operatorname{sech(t)}\end{cases}$$ Is there a way to switch from these equations to the equations: $$\begin{cases}x(\theta)=a\{\ln[\tan(\theta/2)]+\cos\theta\}\\y(\theta)=a\sin(\theta)\end{cases}$$ which depict the same curve?

See http://mathworld.wolfram.com/Tractrix.html for more info. I cannot figure out how they've obtained the last parametric equations I wrote above: I'm not convinced about their result even if it's correct; Once solved for $t$:$$\theta(t)=-\tan^{-1}\operatorname{csch(t)}$$ I plugged back in the first parametric equations I wrote (as they say) and did not obtain the same result. Could one show me the way to get those parametric equations?

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Hint: Ask yourself what the value domain of $\text{sech }t$ is. Then ask yourself what the value domain of $\sin\theta$ is. Are they the same for $\theta>0$ ? If so, then you are half done. The other half consists in verifying that both expressions for x and y satisfy the same functional relation. For the first system, use $\tanh t=\dfrac{\sinh t}{\cosh t}$ and $\cosh^2t-\sinh^2t=1$. For the second, use the Weierstrass substitution.

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