1.

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 `cosec^(2)angleCAB-1=tan^(2)angleABC`.

Answer» 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^(@)`.
`cosec^(2)angleCAB-1`
`=cot^(2)angleCAB`
`=cot^(2)(90^(@)-angleABC)`
`=tan^(2)angleABC`.
Hence `cosec^(2)angleCAB-1=tan^(2)angleABC`.


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