Since (r ⇒ (p ∧ q)) ⇔ ((r ⇒ p) ∧ (r ⇒ q)) is a biconditional (i.e., equivalence), one way is to rewrite (r ⇒ (p ∧ q)) until you get ((r ⇒ p) ∧ (r ⇒ q)), or vice versa. This is because A ⇔ A is a tautology for any formula A. Alternatively, which also works not only for biconditionals, you can rewrite the formula until you get TRUE.