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sin²θ×secθ×cscθ=tanθ
So far I have 1cos²θ/1 × 1/cosθ × 1/sinθ = 1cos²θ/cosθ×sinθ = 1cosθ/sinθ
But I don't know how to get rid of the 1 and then invert that from cot to tan.
 one year ago
 one year ago
sin²θ×secθ×cscθ=tanθ So far I have 1cos²θ/1 × 1/cosθ × 1/sinθ = 1cos²θ/cosθ×sinθ = 1cosθ/sinθ But I don't know how to get rid of the 1 and then invert that from cot to tan.
 one year ago
 one year ago

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nubeerBest ResponseYou've already chosen the best response.0
dw:1354252829403:dw do you get this?
 one year ago

nbedwards934Best ResponseYou've already chosen the best response.0
I changed the sin²θ to 1cos²θ. I'm guessing that's my problem?
 one year ago

nbedwards934Best ResponseYou've already chosen the best response.0
I have in my question what I've worked on so far
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
yes .. don't change. it just leave it like that so u can cancel sin in numerator by the one in denominator.
 one year ago

nbedwards934Best ResponseYou've already chosen the best response.0
wow. that's much easier than I was trying to make it. Thank you
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
yes .. you are welcome :)
 one year ago

nbedwards934Best ResponseYou've already chosen the best response.0
I'll be asking for help on a few other questions like this soon. These are really confusing...
 one year ago
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