AOB is a diameter of a circle with centre O and C is any point on the circle, joining A, C, B, C, and O, C, prove that
`tanangleABC=cotangleACO`

Explanation

since AOB is a diameter and C is any point on the circle,
`:.angleACB` is a semicircular angle.
`:.angleACB=90^(@)`
`:.` AB is a hypotenuse of the right-angled triangle ABC.
Again, since `angleACB=90^(@),:.angleBAC+angleCBA=90^(@)`.
Now, (a) `tanangleABC=tan(90^(@)-angleCAB)`
`=cotangleCAB..................(1)`
Again, in `DeltaAOC,OA=OC[:'` radii of same circle]
`impliesangleOAC=angleACO`
`impliesangleCAB=angleACO`
`:.` from (1) we get, `tanangleABC=cotangleACO`.