# Trigonometric Identity

1. Oct 27, 2008

### brianlane24

cosx/1-sinx + 1-sinx/cosx = 2secx + 2tanx

I can get to 2secx + tanx - tanx, any help is appreciated

2. Oct 27, 2008

### rock.freak667

Type out all the working you have done, if you do not do that, then we won't know where you've reached and how to help you go towards the end result of the proof.

3. Oct 27, 2008

### brianlane24

(cosx)/(1-sinx) x (1+sinx)/(1+sinx) + (1-sinx)/(cosx) x (cosx)/(cosx)
(cosx+sinxcosx)/(1-sin^2x) + (cosx-sinxcosx)/cos^2x
(cosx/cos^2x)+(sinxcosx/cos^2x) + cosx/cos^2x - (sinxcosx/cos^2x)
1/cosx + sinx/cosx + 1/cosx - sinx/cosx
2secx

4. Oct 27, 2008

### rock.freak667

I've tried both sides and can only get 2secx on the LHS and 2cosx/(1-sinx) on the RHS....the problem is written down correctly right?

5. Oct 27, 2008

### brianlane24

I'm fairly certain, that is what the worksheet said,

6. Oct 27, 2008

### brianlane24

When I worked from the RHS, i got to
(2cosx/1-sin^2x) + (2sinxcosx/1-sin^2x)

7. Oct 27, 2008

### brianlane24

Nevermind, thank you, my teacher wrote down the wrong question

8. Oct 27, 2008

### Staff: Mentor

It's no wonder you can't prove it: The equation you gave is not an identity. I tried it with a specific value of x, pi/4, for which sin(pi/4) = sqrt(2)/2 = cos(pi/4), and tan(pi/4) = 1.

The value on the left side was 2sqrt(2), and on the right it was 2sqrt(2) + 2.

Are you sure that:
1. you copied the equation correctly?
2. you weren't supposed to solve the equation rather than prove it was an identity?