Answer
\[\cosh 2x=\cosh^2 x+\sinh^2 x\]
Work Step by Step
To show that:-\[\cosh 2x=\cosh^2 x+\sinh^2 x\]
We know that \[\sinh x=\frac{e^x-e^{-x}}{2}\;\;\;...(1)\]
and \[\cosh x=\frac{e^x+e^{-x}}{2}\;\;\;...(2)\]
Using (2)
\[\cosh 2x=\frac{e^{2x}+e^{-2x}}{2}\;\;\;...(3)\]
Consider \[I=\cosh^2 x+\sinh^2 x\]
Using (1) and (2)
\[I=\left(\frac{e^x+e^{-x}}{2}\right)^2+\left(\frac{e^x-e^{-x}}{2}\right)^2\]
\[I=\left(\frac{e^{2x}+e^{-2x}+2}{4}\right)+\left(\frac{e^{2x}+e^{-2x}-2}{4}\right)\]
\[\Rightarrow I=\frac{2(e^{2x}+e^{-2x})}{4}\]
\[\Rightarrow I=\frac{e^{2x}+e^{-2x}}{2}\;\;\;...(4)\]
Using (3) and (4)
\[\cosh 2x=I\]
Hence ,
\[\cosh 2x=\cosh^2 x+\sinh^2 x\]
