Now, $\text {L.H.S:} \tanh^{-1} \sin \theta = \dfrac {1}{2} \log \left ( \dfrac {1+\sin \theta}{1-\sin \theta} \right ) \cdots \ \cdots (1)$

and $\text {R.H.S} \therefore \cosh^{-1} sec \theta$

$=\log \left(\sec \theta + \sqrt{-1+\sec^2 \theta} \right )$

$= \log (\sec \theta + \tan \theta) = \log \dfrac {1+\sin \theta}{\cos \theta}$

$ =\dfrac {1}{2}\log \left ( \dfrac {1+\sin \theta}{\cos \theta} \right )^2 = \dfrac {1}{2} \log \left ( \dfrac {(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)} \right )$

$ =\dfrac {1}{2}\log \left (\dfrac {1+\sin \theta}{1-\sin \theta} \right )\cdots \ \cdots (2)$

From (1) and (2)

$\tanh^{-1} \sin\theta = \cosh^{-1}\sec\theta$