Need help w/ trigonometry problem: x pos., acute, cos(x)= a, then tan(x)=...?

[TheGreatBuBBy]

if x is a positive acute angle and cos(x)= a, an expression for tan(x) in terms of a is

1) (1-a) / a
2) √(1-(a^2))
3)√ (1-(a^2)) / a
4) 1 / (1-a)

I have been told two different answers (1 and 3). If someone could give me the answer with an explanation, I would appreciate it.

 

[Deleted member 4993]

if x is a positive acute angle and cos(x)= a, an expression for tan(x) in terms of a is

1) (1-a) / a
2) √(1-(a^2))
3)√ (1-(a^2)) / a
4) 1 / (1-a)

I have been told two different answers (1 and 3). If someone could give me the answer with an explanation, I would appreciate it.

What are your thoughts?

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[Steven G]

if x is a positive acute angle and cos(x)= a, an expression for tan(x) in terms of a is

1) (1-a) / a
2) √(1-(a^2))
3)√ (1-(a^2)) / a
4) 1 / (1-a)

I have been told two different answers (1 and 3). If someone could give me the answer with an explanation, I would appreciate it.

Well did you even check to see if 1 and 3 yield the same answers for different a's.

You came here for help, which is good. But what have you done? Exactly where do you need help?

Do you know the definition of cos(x). Given that you have cos(x)=a, can you draw and label the sides of the triangle? Can you then use this triangle to determine tan(x)? If not, then tell us where you are stuck.

 

[TheGreatBuBBy]

I know tan(x) = sin(x)/cos(x)
and cos(x) = a
so tan(x) = sin(x)/a

sin(x) = cos(90-x)

cos(90-x)/a

and now im stuck here.

I dont think it is 1) because even though cos(90) = 1 and cos(90) - cos(x) = 1-a, cos(90-x) and cos(90) - cos(x) are not the same.