## anonymous 5 years ago simple trig identity! prove sinx/1+cosx= 1/sinx - 1/tan x

1. anonymous

so i think you write not right equal

2. anonymous

3. anonymous

honestly, i would if i knew how. i have 2 assignments of an online functions course left to do and once i'm done, i never have to do math again, so all i really need to do is finish them. i've done what i can on my own and now i'm asking for help.

4. anonymous

Note $1-\cos^2x = \sin^2x$ And think about what you can multiply the top and bottom of the LHS (or RHS) to use this fact.

5. anonymous

tan(x)= sin(x)/cos(x)

6. anonymous

My method is more interesting ¬_¬

7. anonymous

$a^2 - b^2 = (a+b)(a-b)$ May also be useful.

8. anonymous

I like Newton's method

9. anonymous

this is all just confusing me more. it's so much easier to understand when i see a solution.

10. anonymous

Sorry, I can't do your problem for you. Ask one of the other people.

11. anonymous

so far i took the right side and got 1/sinx-cosx/sinx which = 1-cosx/sinx but i dont know if that is correct, or what to do next

12. anonymous

That is correct, and useful. How can you get from that to the LHS?

13. anonymous

that's what i don't know.

14. anonymous

i some how need to flip it and chance the - to a +

15. anonymous

Not technically true. Because on the LHS, the sin(X) is on the top, but on the RHS, the sin(x) is on the bottom. See my above advice.

16. anonymous

but there are no ^2 in this proof

17. anonymous

OK. So if you take your LHS and RHS in the new form you have. $\frac{\sin x}{1+\cos x} = \frac{1-\cos x}{\sin x}$ Multiplying both sides by (1+cosx)(sinx) gives a result that is clearly true. (s^2 - 1-c^2) HOWEVER it is best to go directly from LHS -> RHS (or vice versa) and not assume the result and work from that. So can you see how something similar could get from one side to the otehr directly?