10) Consider the following single-qubit rotation gates: 1 point COS -i sin COS sin e-10/2 0 Rx (0) = Ry (0) = Rz(0) = -i sin 0 The Hadamard gate can be realized COS sin COS eie/2 as a sequence of gates of the form H = eid/2 R. (a) R.. (0) R. (B). Identify the values of o, o, 0 and B. O 0 = /4, a = B = 1, 0 = 1/2. O 0 =0, a = 8 = 1/2, 0 = 1/4. O 0 = 1/4, a = B = 1/2, 0 = 1/4. 0 = 1/2, a = B =0 = 1/2