Sin x/1-cosx=1+cosx/sinx
I need to show that one equals the other by choosing one side to prove.
Hello and welcome, I wish to offer two friendly pieces of advice for future topics since you are new here:
i) This is a trigonometry problem, so it should be posted in the trigonometry forum. This forum is for new members to introduce themselves. Posting your questions in the appropriate forum will help ensure you get prompt help and makes things easier for the staff. If you are unsure which forum is most appropriate, use the report post feature to alert the moderators that you wish for your topic to be placed in the best forum.
ii) Use bracketing characters to make clear what the expressions are. This eliminates guesswork on the part of the contributors. In this problem, it is fairly easy to determine what is meant, but sometimes I have seen the lack of bracketing characters result in a waste of time for people as they try to help with the "wrong problem."
Now, we are given to prove:
$\displaystyle \frac{\sin(x)}{1-\cos(x)}=\frac{1+\cos(x)}{\sin(x)}$
I like to begin with the left side, and manipulate it so that the right side results. Look at what the left side is and what we want the result to be. To get the numerator to be $\displaystyle 1+\cos(x)$, a good place to start is to multiply the left side by 1 in the form $\displaystyle \frac{1+\cos(x)}{1+\cos(x)}$:
$\displaystyle \frac{\sin(x)}{1-\cos(x)} \cdot\frac{1+\cos(x)}{1+\cos(x)}$
What does this give you?