Let us suppose that there is a circle having centre O.

Let P is an external point from which two tangents PA and PB are drawn to the circle.

These tangents touch the circle at A and B. 

Now from the figure,

since OA is perpendicular to PA

So ∠OAP = 90

Again from the figure,

since OB is perpendicular to PB

So ∠OBP = 90

Now in the quadrilateral OAPB,

Sum of all interior angles = 360

=> ∠OAP + ∠APB + ∠PBO + ∠BOA = 360

=> 90 + ∠APB + 90 + ∠BOA = 360

=> ∠APB + ∠BOA +  180 = 360

=> ∠APB + ∠BOA = 360 - 180

=> ∠APB + ∠BOA = 180

Hense the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.