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