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Mathematics 26 Online
OpenStudy (anonymous):

Verify each identity:

OpenStudy (anonymous):

\[\frac{ \sin \theta }{ 1-\cot \theta }-\frac{ \cos \theta }{ \tan \theta -1 }= \sin \theta + \cos \theta\]

OpenStudy (anonymous):

i would start by changing the cot. in terms of tangent getting the denominator with tan instead of cot

OpenStudy (anonymous):

nevermind. after some work that doesn't do any good.

OpenStudy (anonymous):

get common denominator first. then all algebra operations. its tedius.

OpenStudy (anonymous):

how would i do that? :)

OpenStudy (anonymous):

(sin(tan-1)/(1-cot)(tan-1))-(cos(1-cot)/(1-cot)(tan-1))= sin+cos

OpenStudy (anonymous):

do it like an algebra problem. forget about the trig functions as something mysterious. they're just expressions

OpenStudy (anonymous):

this problem is so long. type out your work step by step and ill tell you if your doing it right. so much to type out

OpenStudy (anonymous):

did you get it? i can maybe attach a pic of my work. but its a lot to look at

OpenStudy (anonymous):

I got it :) thanks!

OpenStudy (anonymous):

looking at it again. i am curious if there is a way to do it working only from the LHS i spent about 3 hours messing with it and couldnt get it equal. have you tried?

OpenStudy (anonymous):

@hartnn maybe you know if there is a way??

hartnn (hartnn):

yeah first convert everything into sin and cos write cot as cos/sin write tan as sin/cos what u get ?

OpenStudy (anonymous):

sin(1-cos/sin)-cos(sin/cos-1)=sin+cos sin-cos-sin+cos=sin+cos sin-sin+cos-cos=sin+cos 0=sin+cos? i must have made a mistake im not seeing it though.

OpenStudy (anonymous):

wait i think i see it. give me a sec

hartnn (hartnn):

but its sin / (1-cot) you seemed to have simplified sin (1-cot)

OpenStudy (anonymous):

im am to 1/sin-cos and i dont remember what is legal at this point.

OpenStudy (anonymous):

ya i realized that. i would have to do the reciprocal

OpenStudy (anonymous):

this time i did sin(1/(sin/sin-cos/sin)-cos(1/(sin/cos-cos/cos)=sin+cos then i got sin/(sin-cos/sin)-cos/(sin-cos/cos) then to sin^2/(sin-cos)-cos^2/(sin-cos) then sin^2-cos^2/(sin-cos). can i factor a sin-cos out of the top and cancel them to get sin+cos?

hartnn (hartnn):

absolutely correct!

OpenStudy (anonymous):

whew! i spent a lot of time making mistakes. good practice though. its bad when your dreaming about trig identities ha

hartnn (hartnn):

lol, but good work though :-)

OpenStudy (anonymous):

thanks for the help. your a heck of a guy. i don't care what Vegeta says about you.

hartnn (hartnn):

haha, i don't care either :P

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