To determine which equation relating the acute angles [tex]\( A \)[/tex] and [tex]\( C \)[/tex] of a right triangle [tex]\( \triangle ABC \)[/tex] is true, let's analyze each given choice using trigonometric identities.
### Choice A: [tex]\(\tan A = \sin A\)[/tex]
The identity for tangent is:
[tex]\[\tan A = \frac{\sin A}{\cos A}\][/tex]
For [tex]\(\tan A\)[/tex] to equal [tex]\(\sin A\)[/tex], we would need:
[tex]\[\frac{\sin A}{\cos A} = \sin A\][/tex]
This simplifies to:
[tex]\[\sin A = \sin A \cdot \cos A\][/tex]
If we divide both sides by [tex]\(\sin A\)[/tex] (assuming [tex]\(\sin A \neq 0\)[/tex]):
[tex]\[1 = \cos A\][/tex]
This implies that [tex]\(\cos A = 1\)[/tex], which is true only for [tex]\(A = 0^\circ\)[/tex], which is not an acute angle. Therefore, this choice is false in all general cases.
### Choice B: [tex]\(\cos A = \sin (90^\circ - A)\)[/tex]
Using the trigonometric identity:
[tex]\[\sin (90^\circ - A) = \cos A\][/tex]
So:
[tex]\[\cos A = \cos A\][/tex]
This is indeed a correct identity. Thus, this choice is true.
### Choice C: [tex]\(\sin C = \frac{\sin A}{\tan C}\)[/tex]
Recall that:
[tex]\[\tan C = \frac{\sin C}{\cos C}\][/tex]
Substituting this into the given equation:
[tex]\[\sin C = \frac{\sin A}{\frac{\sin C}{\cos C}} = \sin A \cdot \frac{\cos C}{\sin C}\][/tex]
This simplifies to:
[tex]\[\sin C = \sin A \cdot \cos C\][/tex]
This form does not hold as a general identity for the acute angles in a right triangle. Therefore, this choice is false.
### Choice D: [tex]\(\cos A = \tan C\)[/tex]
Using [tex]\(\tan C = \frac{\sin C}{\cos C}\)[/tex] again, we cannot equate [tex]\(\cos A\)[/tex] directly to [tex]\(\tan C\)[/tex] since [tex]\(\cos A\)[/tex] would not generally simplify to this expression. Therefore, this choice is false.
### Choice E: [tex]\(\sin C = \frac{\cos (40^\circ - C)}{\cos A}\)[/tex]
This equation does not align with fundamental trigonometric identities, and there is no standard identity that transforms in this manner. Therefore, this choice is false.
So, after considering all options, the true equation relating the acute angles [tex]\( A \)[/tex] and [tex]\( C \)[/tex] of a right triangle [tex]\( \triangle ABC \)[/tex] is:
B. [tex]\(\cos A = \sin(90^\circ - A)\)[/tex]
The correct answer is choice B.
