Proving a really frustrating trig identity

[Student11]

Hi there.
So I have been trying all afternoon to prove this trig identity. I know it works. I have tried using the classic method of converting everything to sine and cosine then factorising, but I always get stuck being unable to create the tans.
If you could take me through the steps to help me prove this identity it would be much appreciated. Thank you. Hopefully it comes out clear enough. I really need it within the next 8hrs if possible.

Prove that 1+cos(x)/ 1−csc(x) + 1−cos(x)/ 1+csc(x) =−2tan(x) ( tan(x)+1)

 

[lookagain]

Prove that 1+cos(x)/ 1−csc(x) + 1−cos(x)/ 1+csc(x) =−2tan(x) ( tan(x)+1)

You are missing grouping symbols. Type it something similar to this:

Prove that [1 + cos(x)]/[1 − csc(x)] + [1 − cos(x)]/[1 + csc(x)] = −2tan(x)[tan(x) + 1]


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Here are some beginning steps. You need to be able to finish this on your own:


\(\displaystyle \dfrac{1 + cos(x)}{1 - \frac{1}{sin(x)}} \ + \ \dfrac{1 - cos(x)}{1 + \frac{1}{sin(x)}}\)



\(\displaystyle \dfrac{sin(x)}{sin(x)}\cdot\dfrac{1 + cos(x)}{1 - \frac{1}{sin(x)}} \ + \ \dfrac{1 - cos(x)}{1 + \frac{1}{sin(x)}}\cdot \dfrac{sin(x)}{sin(x)}\)



\(\displaystyle \dfrac{sin(x)[1 + cos(x)]}{sin(x) - 1} \ + \ \dfrac{sin(x)[1 - cos(x)]}{sin(x) + 1}\)


\(\displaystyle \dfrac{sin(x) + 1}{sin(x) + 1}\cdot\dfrac{sin(x)[1 + cos(x)]}{sin(x) - 1} \ + \ \dfrac{sin(x)[1 - cos(x)]}{sin(x) + 1}\cdot \dfrac{sin(x) - 1}{sin(x) - 1}\)

 

[mmm4444bot]

Hi Student11:

Please take time to read the forum guidelines; here's a link to the summary page.

(You'll find links to the complete rules and guidelines appear near the bottom.)

Then show us what you've done, up to where you think you're stuck.

Thank you! :cool: