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Prove the following: \[\dfrac{1}{1- \dfrac{1}{1-\dfrac{1}{1-\cos h^2x}}} = \cos h^2 x\]
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$\begin{equation} \text{L.H.S.}= \cfrac{1}{1- \cfrac{1}{1- \cfrac{1}{1- {\cosh^2 x} } } } \end{equation}$

$\begin{equation} = \cfrac{1}{1- \cfrac{1}{1- \cfrac{1}{1-\sinh^2x } } } \end{equation}$

$\begin{equation} = \cfrac{1}{1-\cfrac{1}{ 1+\text{cosech}^2x} } \end{equation}$

$\begin{equation} = \cfrac{1}{1-\cfrac{1}{{\coth^2x} } } \end{equation}$

$\begin{equation} =\dfrac{1}{1-\tanh^2 x } \end{equation}$

$=\dfrac {1}{\text{sech}^2x}$

$=\cosh^2x=\text{R.H.S.}$

$\text{Hence, } \begin{equation} \cfrac{1}{1-\cfrac{1}{1- \cfrac{1}{1-\cos h^2x} } } \end{equation} = \cos h^2x$

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