Question

I have to prove sec(u)-tan(u)=1-sin(u)/cos(u).
So far I've gotten [1/sin(u)]-[sin(u)/cos(u)]= [1-sin(u)/cos(u)].

Answer 1

\[\sec(u) - \tan(u) = \dfrac{1 - \sin(u)}{\cos(u)}\] ?

Answer 2

hint: \(\bf sec(\theta)=\cfrac{1}{cos(\theta)}\qquad tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\)

do the left side

Answer 3

Actually, @alyssa8 seems to have already demonstrated the steps:

\[\sec(u) - \tan(u) = \dfrac{1}{\cos(u)} - \frac{\sin(u)}{\cos(u)}=\dfrac{1 - \sin(u)}{\cos(u)}\]

Answer 4

Yes, that's what I've done so far. Hopefully I'm accurate in those steps. Unless I'm wrong, in which case, I hope you wouldn't mind guiding me.

Answer 5

You basically already proved what needed to be proved which is to start with the LHS and end up with the RHS

Answer 6

Have I really? I thought I had to change it to be exactly like the RHS.

Answer 7

\[\text{LHS} = \sec(u) - \tan(u) = \dfrac{1}{\cos(u)} - \frac{\sin(u)}{\cos(u)}=\dfrac{1 - \sin(u)}{\cos(u)} = \text{RHS}\] Is there something you're confused about here?

Answer 8

I'm terribly sorry. I mis-typed when I asked my question. The proof is sec(u)-tan(u)=cos(u)/1+sin(u). I'm currently to the conclusion of 1-sin(u)/cos(u).

Answer 9

I see.

Answer 10

Close this question and re-post again.

Answer 11

Okay. Thank you.