Concept:The principal value of an inverse trigonometric function must lie within its defined range.Explanation:We evaluate each option using 57π=π+52π.For option A: sin(π+52π)=−sin52π.Since sin−1 has principal range [−2π,2π], and −sin52π<0, the angle is −52π because 52π∈[0,2π].Thus, sin−1[sin(57π)]=−52π.For option B: tan(π+52π)=tan52π, whose principal value is 52π.For option C: cos(π+52π)=−cos52π, giving principal value π−52π=53π.For option D: sec(π+52π)=−sec52π, leading to principal value 53π. Only option A yields −52π.Answer:A. sin−1[sin(57π)]