y=cot[Arccos(3/5)]. Set x=Arccos(3/5), so cosx=3/5.
Hence,
y=cot[Arccos(3/5)]
=cotx
=cosx/sinx
=cosx/Sqrt[1-(cosx)^2]
=(3/5)/Sqrt[1-(3/5)^2]
=(3/5)/Sqrt(1-9/25)
=(3/5)/Sqrt(16/25)
=(3/5)/(4/5)
=3/5*5/4
=3/4
arcocs = cos^(-1). a quanto sarà uguale cotangente (arccos 3/5) = ?