Answer
\[\coth^2 x-1=\csc h^2x\]
Work Step by Step
To show that:-\[\coth^2 x-1=\csc h^2 x\]
We know that \[\sinh x=\frac{e^x-e^{-x}}{2}\;\;\;...(1)\]
and \[\cosh x=\frac{e^x+e^{-x}}{2}\;\;\;...(2)\]
\[\csc h x=\frac{1}{\sinh x}\]
Using (1)
\[\Rightarrow \csc h x=\frac{2}{e^x-e^{-x}}\;\;\;...(3)\]
\[\coth x=\frac{\cosh x}{\sinh x}\]
Using (1) and (2)
\[\coth x=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\]
\[\Rightarrow \coth^2 x=\left(\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\right)^2\]
\[\Rightarrow \coth^2 x-1=\left(\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\right)^2-1\]
\[\Rightarrow \coth^2 x-1=\frac{(e^{x}+e^{-x})^2-(e^x-e^{-x})^2}{(e^{x}-e^{-x})^2}\]
\[\Rightarrow \coth^2 x-1=\frac{2^2}{(e^{x}-e^{-x})^2}\]
Using (3)
\[\Rightarrow \coth^2 x-1=\csc h^2x\]
Hence, \[\coth^2 x-1=\csc h^2x\]
