Let f(x) = x - sin x Now, differentiating both sides w.r.t x ⇒ f’(x) = 1 - cos x We know, -1 ≤ cos x ≤ 1 for all x > 0 ⇒ 1 ≥ -cos x ≥ -1 ⇒ 2 ≥ 1 - cos x ≥ 0 ∴ f’(x) ≥ 0 for all x > 0 Hence, f(x) is an increasing function for all x > 0. And since, f(x) is an increasing function for all x > 0, then x > sin x for all x > 0. Hence, both statement I & II are correct and statement II is correct explanation of statement I.