If A, B and C be the angles of a triangle, then out of the following, the incorrect relation is:

If A, B and C be the angles of a triangle, then $$\angle$$ A + $$\angle$$ B + $$\angle$$ C = $$180^\circ$$

=> $$\angle$$ A + $$\angle$$ B = $$180^\circ$$ - $$\angle$$ C

=> $$\frac{\angle A+\angle B}{2}=90^\circ-\frac{\angle C}{2}$$ -------------(i)

(A) : $$tan(\frac{A+B}{2})=tan(90^\circ-\frac{\angle C}{2})=cot(\frac{\angle C}{2})\neq sec(\frac{\angle C}{2})$$

(B) : $$cot(\frac{A+B}{2})=cot(90^\circ-\frac{\angle C}{2})=tan(\frac{\angle C}{2})$$

(C) : $$sin(\frac{A+B}{2})=sin(90^\circ-\frac{\angle C}{2})=cos(\frac{\angle C}{2})$$

(D) : $$cos(\frac{A+B}{2})=cos(90^\circ-\frac{\angle C}{2})=sin(\frac{\angle C}{2})$$

=> Ans - (A)