Which trigonometric identity follows directly by dividing $AB^2+BC^2=AC^2$ by $AC^2$ in a right triangle? 

Principle: start from the Pythagoras theorem in a right triangle and convert side ratios into trigonometric ratios. Since $AB^2+BC^2=AC^2$, dividing every term by $AC^2$ gives $$ \frac{AB^2}{AC^2}+\frac{BC^2}{AC^2}=1. $$ Now $\frac{AB}{AC}=\cos A$ and $\frac{BC}{AC}=\sin A$, so this becomes $$ \cos^2 A+\sin^2 A=1. $$ Option A comes from dividing by $AB^2$, not $AC^2$. Option C comes from dividing by $BC^2$. Option D is not a standard identity and is generally false.