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Babynini
 one year ago
Prove the identity
(tanx)/(1cosx) = cscx(1+secx)
Babynini
 one year ago
Prove the identity (tanx)/(1cosx) = cscx(1+secx)

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.0I've gotten up to (sinx/cosx)(1/(1cosx)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0is that correct so far?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok and you can write that as:\[\frac{\frac{\sin(x)}{\cos(x)}}{1\cos(x)} \\ \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1\cos(x)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oh that is what you wrote

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1\cos(x)} \cdot \frac{1+\cos(x)}{1+\cos(x)}\] see what this does maybe

freckles
 one year ago
Best ResponseYou've already chosen the best response.1multiply the second and last fraction

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0so the last fraction is just like multiplying by 1 right?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0It'd be (sinx/cosx)((1+cosx)/(1cos^2x))

freckles
 one year ago
Best ResponseYou've already chosen the best response.1right and 1cos^2(x) is sin^2(x)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\sin(x)}{\cos(x)} \cdot \frac{1+\cos(x)}{\sin^2(x)} \\ \frac{\sin(x)}{\sin^2(x)} \frac{1+\cos(x)}{\cos(x)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1can you see where to go from here?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\sin(x)}{\sin^2(x)}(\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)})\] how about here?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1cos(x)/cos(x)=? 1/cos(x)=? sin(x)/sin^2(x)=?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.01) = 1 2) = secx 3) I'm not sure about that one :P

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0where did you get the cosx/cosx?

freckles
 one year ago
Best ResponseYou've already chosen the best response.11) good 2) good 3) do you know how to simplify say something like u/u^2 ? separated the fraction to "where did you get the cosx/cosx? " \[\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c} \\ \text{ we had } \frac{1+\cos(x)}{\cos(x)}=\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{u}{u^2}=\frac{u}{u \cdot u}=\frac{\cancel{u}}{\cancel{u} u}=\frac{1}{u} \\ \text{ or just use law of exponents } \\ \frac{u}{u^2}=u^{12}=u^{1}=\frac{1}{u} \\ \frac{\sin(x)}{\sin^2(x)}=\frac{1}{\sin(x)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1What we did: \[\frac{\tan(x)}{1\cos(x)} \\ \tan(x) \cdot \frac{1}{1\cos(x)} \\ \frac{\sin(x)}{\cos(x)} \frac{1}{1\cos(x)} \\ \frac{\sin(x)}{\cos(x)} (\frac{1}{1\cos(x) } \cdot 1) \\ \frac{\sin(x)}{\cos(x)} (\frac{1}{1\cos(x) } \cdot \frac{1+\cos(x)}{1+\cos(x)} ) \\ \frac{\sin(x)}{\cos(x)}(\frac{1+\cos(x)}{1\cos^2(x)}) \\ \frac{\sin(x)}{\cos(x)} \frac{1+\cos(x)}{\sin^2(x)} \\ \frac{\sin(x)}{\sin^2(x)} \frac{1+\cos(x)}{\cos(x)} \\ \frac{1}{\sin(x)} \frac{1+\cos(x)}{\cos(x)} \\ \frac{1}{\sin(x)} (\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)}) \\ \csc(x)(\sec(x)+1) \\ \csc(x)(1+\sec(x))\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0aw thanks for taking the time to type that all up!

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0haha I guess I should do a few more practice ones. Sigh. Thanks so much for the help :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1np trig identities can be tricky (Takes practice practice) but fun just think you are doing a puzzle or something puzzles should be fun

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Tricky may be an understatement xP haha sure sure.
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