Arcos(-√3/2)+arcctg(1/√3) 1)п/6 2)2п/3 3)7п/6 4)-п/6

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вопрос

The sum of the arcsine of -√3/2 and the arccotangent of 1/√3 can be simplified as follows:

1) Arcos(-√3/2) = π/6
2) Arcctg(1/√3) = arctan(√3) = π/3

Therefore, Arcos(-√3/2) + Arcctg(1/√3) = π/6 + π/3 = 2π/6 + 4π/6 = 6π/6 = π

So the correct answer is 1) π/6