Démontrer que tg A = sin A/cos A

Dans un triangle ABC rectangle en B, on a : tg $\widehat{A}$ = $\frac{sin \widehat{A}} {cos \widehat{A}}$ .

On sait que : tg $\widehat{A}$ = $\frac{C\hat{o} t\acute {e} oppos\acute {e}}{C\hat{o}t\acute {e} adjacent}$ = $\frac{BC}{AB}$ .
cos $\widehat{A}$ = $\frac{C\hat{o}t\acute {e} adjacent}{Hypot\acute {e}nuse}$ = $\frac{AB}{AC}$
sin $\widehat{A}$ = $\frac{C\hat{o}t\acute {e} oppos\acute {e}}{Hypot\acute {e}nuse}$ = $\frac{BC}{AC}$
$\frac{sin \widehat{A}}{cos \widehat{A}}$ = $\frac{{BC}{AC}}{{AB}{AC}}$ = $\frac{{BC \times AC}}{{AB \times AC}}$ = $\frac{BC}{AB}$ = tg $\widehat{A}$