Mark M. answered 10/06/23
Retired math prof. Very extensive Precalculus tutoring experience.
cscθ - cotθ = 1/sinθ - cosθ / sinθ = (1 - cosθ) / sinθ
So, y = ln l cscθ - cotθ l = ln l (1 - cosθ) / sinθ l = ln l 1 - cosθ l - ln l sinθ l
y' = (sinθ) / (1 - cosθ) - cosθ / sinθ = (sin2θ - cosθ + cos2θ) / [sinθ(1 - cosθ)] = (1 - cosθ) / [sinθ(1 - cosθ)]
So, y' = 1 / sinθ = (sinθ)-1
Therefore, y" = -(sinθ)-2(cosθ) = -cosθ / sin2θ = -(1 / sinθ)(cosθ / sinθ) = -cscθcotθ