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