simple trig identity!
prove
sinx/1+cosx= 1/sinx - 1/tan x

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- anonymous

simple trig identity!
prove
sinx/1+cosx= 1/sinx - 1/tan x

- katieb

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- anonymous

so i think you write not right equal

- anonymous

you should do this on your own. it will help you learn about manipulation of trigonometric identities.

- 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.

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## More answers

- 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.

- anonymous

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

- anonymous

My method is more interesting ¬_¬

- anonymous

\[a^2 - b^2 = (a+b)(a-b) \]
May also be useful.

- anonymous

I like Newton's method

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

that's what i don't know.

- anonymous

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

- 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.

- anonymous

but there are no ^2 in this proof

- 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?

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