4. (a) One may avoid the absolute values that appear in the derivative of arcesc(r) by changing the definition of this functions. In defining arcesc(x), we restricted the domain of sin(2) to [-},=] in order to ensure injectivity. Alternatively, we could have restricted the domain of sin(r) to (0, 7/2] U (#, 3x/2], where it is still injective. Prove that if we define arcesc(r)using the new choice of domain: 3TT arcesc : (-co, -1] U [1, co) - (0, ;]U(#, 2 1 then the derivative is given by # arcesc() = _V72.(b) The following "theorem" is not quite true as stated: Flawed "Theorem": For the function arctan : R - (-$, ;), for every r, y e R we have rty arctan(r) + arctan(y) = arctan( 1 - cy Fake "Proof": Proof. Let 61 = arctan(a) and 02 = arctan(y). Then we get tan(61 ) = r and tan(62) = y. tan(61)+tan(62) Now, tan(01 + 02) = 1-tan(1) tan(82) I-ry 01 + 02 = arctan(). Thus, arctan(r) + arctan(y) = arctan( Explain the problem with the statement of the theorem and the errors in the proof. Then fix them: correct the statement, and write a correct proof