Standard transform pair: L{sinh(at)} = a/(s²−a²). Therefore L{(1/a) sinh(at)} = (1/a)·a/(s²−a²) = 1/(s²−a²). So the inverse Laplace transform of 1/(s²−a²) is a⁻¹ sinh(at), matching option (D). Option (C), 'sinh at' without the 1/a factor, is the classic distractor obtained by forgetting to divide by a.
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