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 2 years 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.
 2 years 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.

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nubeer
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1354252829403:dw do you get this?

nbedwards934
 2 years ago
Best ResponseYou've already chosen the best response.0I changed the sin²θ to 1cos²θ. I'm guessing that's my problem?

nbedwards934
 2 years ago
Best ResponseYou've already chosen the best response.0I have in my question what I've worked on so far

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

nbedwards934
 2 years ago
Best ResponseYou've already chosen the best response.0wow. that's much easier than I was trying to make it. Thank you

nubeer
 2 years ago
Best ResponseYou've already chosen the best response.0yes .. you are welcome :)

nbedwards934
 2 years ago
Best ResponseYou've already chosen the best response.0I'll be asking for help on a few other questions like this soon. These are really confusing...
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