First, note that [tex]m\angle A+m\angle C=90^{\circ}.[/tex] Then
[tex]m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.[/tex]
Consider all options:
A.
[tex]\tan A=\dfrac{\sin A}{\sin C}[/tex]
By the definition,
[tex]\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.[/tex]
Now
[tex]\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.[/tex]
Option A is true.
B.
[tex]\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.[/tex]
By the definition,
[tex]\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.[/tex]
Then
[tex]\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.[/tex]
Option B is false.
3.
[tex]\sin C = \dfrac{\cos A}{\tan C}.[/tex]
By the definition,
[tex]\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.[/tex]
Now
[tex]\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.[/tex]
Option C is false.
D.
[tex]\cos A=\tan C.[/tex]
By the definition,
[tex]\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.[/tex]
As you can see [tex]\cos A\neq \tan C[/tex] and option D is not true.
E.
[tex]\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.[/tex]
By the definition,
[tex]\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.[/tex]
Then
[tex]\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.[/tex]
This option is false.
