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The depth of a lake at a certain point, wich is a function of the distance of that point from shore, is 30 feet.

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) = .

(01.04 lc)multiply 1 over 6 multiplied by negative 1 over 2. negative 1 over 3 negative 1 over 12 1 over 12 1 over 3