`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
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