Using the definitions of hyperbolic functions in terms of exponentials show that sech^2(x) = 1-tanh^2(x)

tanh(x) = ((e^x-e^-x)/2)/((e^x+e^-x)/2) 1 - tanh²(x) = 1-((e^x-e^-x)/(e^x+e^-x))²  = ((e^2x+e^-2x+2)-(e^2x+e^-2x-2))/(e^x+e^-x)² = (2e^x.2e^-x)/(e^x+e^-x)² = 4/(e^x+e^-x)² = sech²x