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I presume that sin-1x is being used to represent the inverse sin function (I prefer arcsin x to avoid possible confusion).

Make use of the trignometirc relationships:

cos2θ + sin2θ = 1

⇒ cosθ = √(1 - sin2θ)

cotθ = cosθ/sinθ

= √(1 - sin2θ)/sinθ

sin(arcsin x) = x

Then:

cot(arcsin(x)) = √(1 - sin2(arcsin(x))/sin(arcsin(x))

= √(1 - x2)/x

⇒ cot(arcsin(2/3)) = √(1 - (2/3)2)/(2/3)

= √(9/32 - 4/32) ÷ 2/3

= √(9 - 4) x 1/3 x 3/2

= 1/2 x √5

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Q: How do you solve cot parenthesis sin to the negative 1 parenthesis 2 over 3 closed parenthesis and another closed parenthesis?
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