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Answer 1 · 2016-02-05T13:51:27

I have to guess here but I'm assuming that you would like to prove the following identity:

$sin x / (1 - cos x) - (1 + cos x )/ sin x = 0$

Let's start by bringing the second fraction to the right-hand side:

$<=> sin x / (1 - cos x) = (1 + cos x)/ sin x$

Now, in order to get rid of the fractions, you should multiply both sides with both denominators:

$<=> sin x * sin x = (1 + cos x) * (1 - cos x)$

... make use of the formula $(a-b)(a+b) = a^2 - b^2$ ...

$<=> sin^2 x = 1^2 - cos^2 x$

... add $cos^2 x$ on both sides...

$<=> sin^2 x + cos^2 x = 1$

However, this is a well-known identity.

As your original equation is equivalent to this identity and the identity certainly holds, your equation holds as well.

q.e.d.