Question
Without using truth table show that : $\sim(p \vee q) \vee(\sim p \wedge q) \equiv \sim p$
~(p v q)v(~p ∧ q)
≡~(p v q)v~(p ∨ ~q) by De Morgan's Law
≡~[(p ∨ q) ∧ (p ∨ ~q)] by De Morgan's Law
≡~{[(p ∨ q) ∧ p] ∨ [(p ∨ q)∧ ~q)]} by Distributive Law
≡ ~{[p] ∨ [(p ∨ q) ∧ ~q]} by Absorption Law
≡ ~{[p] ∨ [(p∧ ~q) ∨ (q ∧ ~q)]} by Distributive Law
≡~{[p] ∨ [(p ∧ ~q) ∨ F]} by Complement Law
≡~{[p] ∨ [(p ∧ ~q)]} by Identity Law
≡~p ∧ (~p ∨ q) by De Morgan's Law
≡ ~p by Absorption Law
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$\cos ^{-1}\left(\frac{3 \cos 3 x-4 \sin 3 x}{5}\right)$