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Show that $\sin^{-1}(\tanh x)=\tan^{-1}(\sinh x)$. Got a hint that $\sin\theta=\tanh x$ but I still don't know how to proceed...

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  • $\begingroup$ Does $\tan hx$ denote $\tan (hx)$ or $(\tan h) \times x$? $\endgroup$ – Paul Oct 3 '14 at 2:07
  • $\begingroup$ @Paul: Neither. $\tanh x$ denotes the hyperbolic tangent of x. $\endgroup$ – Lucian Oct 3 '14 at 3:13
  • $\begingroup$ @Lucian Thanks. I see! $\endgroup$ – Paul Oct 3 '14 at 3:49
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Differentiate both to show they share the same derivative. Then, check one value to show equality is met. (if $f'(x)=g'(x)$ on a connected domain then $f(x)=g(x)+c$. You want $c=0$)

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Let $\sin\theta=\tanh x$. $\sin^{-1}(\tanh x)=\sin^{-1}(\sin\theta)=\theta$.

We only need to prove $$\sinh x= \tan \theta. $$ To prove it, by using $\sin\theta=\tanh x$, it only need to prove that

$$\cos \theta =\frac{1} {\cosh x}.$$

In fact: $$\cos^2 \theta=1-\sin^2 \theta=(1-\sin \theta)(1+\sin \theta)=\frac{4}{(e^x+e^{-x})^2}.$$

Now it is easy to get our conclusion.

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