Babynini
  • Babynini
Prove the identity (tanx)/(1-cosx) = cscx(1+secx)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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Babynini
  • Babynini
I've gotten up to (sinx/cosx)(1/(1-cosx)
Babynini
  • Babynini
is that correct so far?
freckles
  • freckles
ok and you can write that as:\[\frac{\frac{\sin(x)}{\cos(x)}}{1-\cos(x)} \\ \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1-\cos(x)}\]

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freckles
  • freckles
oh that is what you wrote
freckles
  • freckles
lol
freckles
  • freckles
\[\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{1-\cos(x)} \cdot \frac{1+\cos(x)}{1+\cos(x)}\] see what this does maybe
freckles
  • freckles
multiply the second and last fraction
Babynini
  • Babynini
so the last fraction is just like multiplying by 1 right?
freckles
  • freckles
yep
Babynini
  • Babynini
It'd be (sinx/cosx)((1+cosx)/(1-cos^2x))
freckles
  • freckles
right and 1-cos^2(x) is sin^2(x)
freckles
  • freckles
\[\frac{\sin(x)}{\cos(x)} \cdot \frac{1+\cos(x)}{\sin^2(x)} \\ \frac{\sin(x)}{\sin^2(x)} \frac{1+\cos(x)}{\cos(x)}\]
freckles
  • freckles
can you see where to go from here?
Babynini
  • Babynini
Emm no haha sorry.
freckles
  • freckles
\[\frac{\sin(x)}{\sin^2(x)}(\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)})\] how about here?
freckles
  • freckles
cos(x)/cos(x)=? 1/cos(x)=? sin(x)/sin^2(x)=?
Babynini
  • Babynini
1) = 1 2) = secx 3) I'm not sure about that one :P
Babynini
  • Babynini
where did you get the cosx/cosx?
freckles
  • freckles
1) 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? " \[\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c} \\ \text{ we had } \frac{1+\cos(x)}{\cos(x)}=\frac{1}{\cos(x)}+\frac{\cos(x)}{\cos(x)}\]
Babynini
  • Babynini
aah I see, ok :)
Babynini
  • Babynini
3) um, i'm not sure.
freckles
  • freckles
\[\frac{u}{u^2}=\frac{u}{u \cdot u}=\frac{\cancel{u}}{\cancel{u} u}=\frac{1}{u} \\ \text{ or just use law of exponents } \\ \frac{u}{u^2}=u^{1-2}=u^{-1}=\frac{1}{u} \\ \frac{\sin(x)}{\sin^2(x)}=\frac{1}{\sin(x)}\]
freckles
  • freckles
and 1/sin(x) is....
Babynini
  • Babynini
cscx!
freckles
  • freckles
yep
freckles
  • freckles
What 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
  • Babynini
aw thanks for taking the time to type that all up!
Babynini
  • Babynini
haha I guess I should do a few more practice ones. Sigh. Thanks so much for the help :)
freckles
  • freckles
np trig identities can be tricky (Takes practice practice) but fun just think you are doing a puzzle or something puzzles should be fun
Babynini
  • Babynini
Tricky may be an understatement xP haha sure sure.

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