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1+((e^x-e^-x)/2)^2 = ((e^x+e^-x)/2)^2

Can someone explain how does the 1st equation is equal to the 2nd one

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  1. LS = 1 + (e^(2x) - 2e^0 + e^(-2x) )/4
    = (4 + e^(2x) - 2 + e^(-2x) )/4
    = ( e^(2x) + 2e^0 + e^(-2x) )/4
    = ( (e^x + e^-x)/2 )^2
    = RS

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