Question

Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

A. tan A = sin A/sin C
B. cos A = tan (90-A)/sin (90-C)
C. sin C = cos A/tan C
D. cos A = tan C
E. sin C = cos (90-C)/tan A

Answer 1

The right anwer is option A.
tan A = sin A / sin C
sin C = sin A / tan A = sin A / (sin A / cos A) = cos A
sin C = cos A

Answer 2

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=(\sin A)/(\sin C)`

By the definition,

`\tan A=(BC)/(AB),\\ \\\sin A=(BC)/(AC),\\ \\\sin C=(AB)/(AC).`

Now

`(\sin A)/(\sin C)=((BC)/(AC))/((AB)/(AC))=(BC)/(AB)=\tan A.`

Option A is true.

B.

`\cos A=(\tan (90^(\circ)-A))/(\sin (90^(\circ)-C)).`

By the definition,

`\cos A=(AB)/(AC),\\ \\\tan (90^(\circ)-A)=(\sin(90^(\circ)-A))/(\cos(90^(\circ)-A))=(\sin C)/(\cos C)=((AB)/(AC))/((BC)/(AC))=(AB)/(BC),\\ \\\sin (90^(\circ)-C)=\sin A=(BC)/(AC).`

Then

`(\tan (90^(\circ)-A))/(\sin (90^(\circ)-C))=((AB)/(BC))/((BC)/(AC))=(AB\cdot AC)/(BC^2)\\eq (AB)/(AC).`

Option B is false.

3.

`\sin C = (\cos A)/(\tan C).`

By the definition,

`\sin C=(AB)/(AC),\\ \\\cos A=(AB)/(AC),\\ \\\tan C=(AB)/(BC).`

Now

`(\cos A)/(\tan C)=((AB)/(AC))/((AB)/(BC))=(BC)/(AC)\\eq \sin C.`

Option C is false.

D.

`\cos A=\tan C.`

By the definition,

`\cos A=(AB)/(AC),\\ \\\tan C=(AB)/(BC).`

As you can see
`\cos A\\eq \tan C` and option D is not true.

E.

`\sin C = (\cos(90^(\circ)-C))/(\tan A).`

By the definition,

`\sin C=(AB)/(AC),\\ \\\cos (90^(\circ)-C)=\cos A=(AB)/(AC),\\ \\\tan A=(BC)/(AB).`

Then

`(\cos(90^(\circ)-C))/(\tan A)=((AB)/(AC))/((BC)/(AB))=(AB^2)/(AC\cdot BC)\\eq \sin C.`

This option is false.