Another trigonometric identity math question...
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Another trigonometric identity math question...

[From: ] [author: ] [Date: 12-04-03] [Hit: ]
Thats identical to the top so it all cancels and becomes 1, leaving just (1 + sin(x)) / cos(x).This is somewhat cheating, but everything I did was valid.I didnt cross the equation sign and I only multiplied by 1 (twice), albeit in a fancy way.......
I've decided I'm not very good with these questions:/
If you could help and show your work it would be greatly appreciated! I just don't know where to begin for this one...

(1 + sinx + cosx) / (1 - sinx + cosx) = (1 + sinx) / cosx

Thanks again!

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Alright, so cross multiplying would verify it, but that's not allowed with identities. So instead I'm going to try to do it all on one side by multiplying and dividing by the right hand side and hoping it all goes to 1 times the stuff on the right.

[(1 + sin(x)) / cos(x)] * [cos(x)(1 + sinx + cosx) / (1 - sinx + cosx)(1 + sin(x))]

Second term bottom:
1 + sin(x) - sin(x) - sin²(x) + cos(x) + sin(x)cos(x)
1 - sin²(x) + cos(x) + sin(x)cos(x)
cos²(x) + cos(x) + sin(x)cos(x)
cos(x)(cos(x) + 1 + sin(x))

That's identical to the top so it all cancels and becomes 1, leaving just (1 + sin(x)) / cos(x).


This is somewhat cheating, but everything I did was valid. I didn't "cross the equation sign" and I only multiplied by 1 (twice), albeit in a fancy way.
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