Trig identities...harder one

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Trig identities...harder one

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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  • hba
Try this. First, change all tan(x) and cot(x) into sin(x)/cos(x) and cos(x)/sin(x) respectively. Then combine the fractions using common denominator which is cos(x)sin(x). -Hba.
  • hba
@AmTran_Bus Try it and tell me what you get ?

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Other answers:

I kept getting it to equal sin2. Let me give it another try.
  • hba
No if you try to reduce it,It will come to \[ \sin^2x -\cos^2x \]
Agreed. I see you now. I have it on my paper. But now you have your +2cos^2 on there. I know sin+cos=1 (as long as they are both squared)...but what about the rest?
  • hba
Then put it in the actual ques. \[\sin^2x- \cos^2x + 2\cos^2x = 1\]
  • hba
Now you know how it goes ? @AmTran_Bus
I understand that, yes. But maybe I just can't see it because I can't carry it on.
  • hba
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  • hba
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Oh yes. Thanks and Happy Thanksgiving.
  • hba
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