(i) A ∪ (A ∩ B) = A
As we know union is distributive over intersection
Therefore, A ∪ (A ∩ B)
(A ∪ A) ∩ (A ∪ B) [Here, A ∪ A = A]
A ∩ (A ∪ B)
A
(ii) A ∩ (A ∪ B) = A
As we know union is distributive over intersection
Therefore, (A ∩ A) ∪ (A ∩ B)
A ∪ (A ∩ B) [Here, A ∩ A = A]
A