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Prove this trig identity csc^4-csc^2=cot^4+cot^2

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look at the LHS...try factoring out csc^2 what do you get?

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not exactly....if you factor out csc^2 then csc^4 becomes csc^2 but csc^2 doesn't become csc^2
Would it even exist, that is, would it cancel? Or just be csc?
what is csc^2/csc^2?
ohhhh. I feel dumb. Got ya.
so what should the factored form be?
now what is csc^2 - 1?
so uou have \[\huge \csc^2 (\cot^2)\] now here comes a tricky your identities and give me the equation for csc^2
csc2 = 1/ also equals cot+1
cot^2 + 1 not cot + 1
Sorry, your exactly right.
you should check your notes and make sure you wrote them might get them wrong... \[\huge \sin^2 + \cos^2 = 1\] \[\huge \tan^2 + 1 = \sec^2\] \[\huge \cot^2 + 1 = \csc^2\]
anyway...back to what i was saying... csc^2 is cot^2 + 1 so \[\huge \csc^2 (\cot^2) \implies (\cot^2 + 1)(\cot^2)\] now expand that
would the lhs say csc cot ^4? The RHS is cot^4 +cot^2
i think you're confused...let me flip it... \[\huge \cot^2 \theta (\cot^2\theta + 1)\] do you know how to distribute that now?
Its not that I'm confused as much as tired my friend. |dw:1353026333309:dw|
well anyway... cot^2(cot^2 + 1) becomes cot^4 + cot^2 so you have just proven the identity

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