Withat are the missing length(s) of the sides of the triangle labeled 'a' and ‘b’.

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .