Respuesta :

Answer:

Option D. [tex]\frac{1}{tanA}[/tex]

Step-by-step explanation:

tan ( 90-A ) = [tex]\frac{sin(90-A)}{cos(90-A)}[/tex]

since  sin (90-A) = cos A

and cos (90-A) = sin A

So  [tex]\frac{sin(90-A)}{cos(90-A)}=\frac{CosA}{SinA}=cot A=\frac{1}{tanA}[/tex]

Option D. [tex]\frac{1}{tanA}[/tex] is the correct answer.

Answer:

d. 1/tanA

Step-by-step explanation:

In order to answer, you have to replace x=90-A in the trigonometric identity to express the tangent in function of sine and cosine.

[tex]tan(90-A)=\frac{sin(90-A)}{cos(90-A)}[/tex]

Now you have to apply the following trigonometric identities:

Sin(α-β)=Sin(α)Cos(β)-Cos(α)Sin(β)

Cos(α-β)=Cos(α)Cos(β)+Sin(α)Sin(β)

In this case, α=90 and β=A

Therefore:

[tex]\frac{Sin(90)Cos(A)-Cos(90)Sin(A)}{Cos(90)Cos(A)+Sin(90)Sin(A)}[/tex]

But Sin(90)=1 and Cos(90)=0

[tex]\frac{Cos(A)}{Sin(A)} = \frac{1}{tan(A)}[/tex]

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