`cosh^2(x) = (1+coshx)/2` Verify the identity.

`cosh^2(x) = (1+coshx)/2`

proof:

LHS=>

`cosh^2(x)=1+sinh^2(x) `

`=1+[(e^x-e^-x)/2]^2 `

`=1+(1/4)[(e^x-e^(-x))]^2] `

`=1+(1/4)[e^(2x)+e^(-2x) -2e^x e^(-x)] `

`=1+(1/4)[e^(2x)+e^(-2x) - 2] `

`=1+(1/4)[2(e^(2x)+e^(-2x))/2 -2]`

on taking 2 common and cancelling, we get

`=1+(1/2)[(e^(2x)+e^(-2x))/2 -1] =1+(1/2)[(e^(2x)+e^(-2x))/2 ] - (1/2)`

`= (1/2)+(1/2)[(e^(2x)+e^(-2x))/2 ]`

`= (1/2)+(1/2)[cosh(2x)]`

`=(1+cosh(2x))/2`

but RHS=>`(1+coshx)/2`

so LHS not equal to RHS