Answer
\[\sinh (x+y)=\sinh x\,\cosh y+\cosh x\,\sinh y\]
Work Step by Step
To show that:-\[\sinh (x+y)=\sinh x\,\cosh y+\cosh x\,\sinh y\]
We know that \[\sinh x=\frac{e^x-e^{-x}}{2}\;\;\;...(1)\]
and \[\cos hx=\frac{e^x+e^{-x}}{2}\;\;\;...(3)\]
Using (1)
\[\Rightarrow \sinh (x+y)=\frac{e^{(x+y)}-e^{-(x+y)}}{2}\]
\[\Rightarrow \sinh (x+y)=\frac{e^xe^y-e^{-x}e^{-y}}{2}\;\;\;...(4)\]
Consider \[I=\sinh x\,\cosh y+\cosh x\,\sinh y\]
Using (2) and (3)
\[I=\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^y+e^{-y}}{2}\right)+\left(\frac{e^x+e^{-x}}{2}\right)\left(\frac{e^y-e^{-y}}{2}\right)\]
\[I=\frac{e^{x}e^{y}+e^{x}e^{-y}-e^{-x}e^{y}-e^{-x}e^{-y}}{4}+\frac{e^{x}e^{y}-e^{x}e^{-y}+e^{-x}e^{y}-e^{-x}e^{-y}}{4}\]
\[I=\frac{2(e^xe^y-e^{-x}e^{-y})}{4}\]
\[\Rightarrow I=\frac{e^xe^y-e^{-x}e^{-y}}{2}\;\;\;...(5)\]
Using (4) and (5)
\[\sinh (x+y)=I\]
Hence , \[\sinh (x+y)=\sinh x\,\cosh y+\cosh x\,\sinh y\]
