`lim_(theta ->0) (cos(theta) - 1)/sin(theta)`
using L'Hopital's rule we get
`lim_(theta ->0) ((cos(theta) - 1)')/(sin(theta)')`
`=lim_(theta ->0) (-sin(theta)/cos(theta))`
as `theta->0` we get
`= (-sin(0)/cos(0)) = 0`
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