Answer
Let $\sin \theta =\tanh x$
$\sin^{-1}(\tanh x)=\sin^{-1}(\tanh x)=\sin^{-1}(\sin\theta)=\theta$
So we only need to prove
$\sinh x=\tan \theta$
To prove it ,by using $\sin\theta=\tanh x$,it only nedd to prove that $
\cos \theta=\frac{1}{\cosh x}$
In fact,$\cos^2\theta=1-\sin^2 \theta=(1-\sin\theta)(1+\sin\theta)=(1-\tanh x)(1+\tanh x)=\frac{4}{(e^x+e^{-x})^2}$
Because $\frac{1}{\cosh^2x}=\frac{4}{(e^x+e^{-x})^2}$
Hence,our conclusion is proved.
Work Step by Step
Let $\sin \theta =\tanh x$
$\sin^{-1}(\tanh x)=\sin^{-1}(\tanh x)=\sin^{-1}(\sin\theta)=\theta$
So we only need to prove
$\sinh x=\tan \theta$
To prove it ,by using $\sin\theta=\tanh x$,it only nedd to prove that $
\cos \theta=\frac{1}{\cosh x}$
In fact,$\cos^2\theta=1-\sin^2 \theta=(1-\sin\theta)(1+\sin\theta)=(1-\tanh x)(1+\tanh x)=\frac{4}{(e^x+e^{-x})^2}$
Because $\frac{1}{\cosh^2x}=\frac{4}{(e^x+e^{-x})^2}$
Hence,our conclusion is proved.
