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AmTran_Bus
 3 years ago
Prove this trig identity
csc^4csc^2=cot^4+cot^2
AmTran_Bus
 3 years ago
Prove this trig identity csc^4csc^2=cot^4+cot^2

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AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1353025115000:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0look at the LHS...try factoring out csc^2 what do you get?

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1353025247991:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not exactly....if you factor out csc^2 then csc^4 becomes csc^2 but csc^2 doesn't become csc^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353025341525:dw

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1Would it even exist, that is, would it cancel? Or just be csc?

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1ohhhh. I feel dumb. Got ya.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what should the factored form be?

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1353025535036:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now what is csc^2  1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so uou have \[\huge \csc^2 (\cot^2)\] now here comes a tricky part...review your identities and give me the equation for csc^2

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1csc2 = 1/sin...it also equals cot+1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cot^2 + 1 not cot + 1

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry, your exactly right.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you should check your notes and make sure you wrote them right....you might get them wrong... \[\huge \sin^2 + \cos^2 = 1\] \[\huge \tan^2 + 1 = \sec^2\] \[\huge \cot^2 + 1 = \csc^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anyway...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

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1would the lhs say csc cot ^4? The RHS is cot^4 +cot^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think you're confused...let me flip it... \[\huge \cot^2 \theta (\cot^2\theta + 1)\] do you know how to distribute that now?

AmTran_Bus
 3 years ago
Best ResponseYou've already chosen the best response.1Its not that I'm confused as much as tired my friend. dw:1353026333309:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well anyway... cot^2(cot^2 + 1) becomes cot^4 + cot^2 so you have just proven the identity
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