Verify each identity:
\[\frac{ \sin \theta }{ 1-\cot \theta }-\frac{ \cos \theta }{ \tan \theta -1 }= \sin \theta + \cos \theta\]
i would start by changing the cot. in terms of tangent getting the denominator with tan instead of cot
nevermind. after some work that doesn't do any good.
get common denominator first. then all algebra operations. its tedius.
how would i do that? :)
(sin(tan-1)/(1-cot)(tan-1))-(cos(1-cot)/(1-cot)(tan-1))= sin+cos
do it like an algebra problem. forget about the trig functions as something mysterious. they're just expressions
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
did you get it? i can maybe attach a pic of my work. but its a lot to look at
I got it :) thanks!
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?
@hartnn maybe you know if there is a way??
yeah first convert everything into sin and cos write cot as cos/sin write tan as sin/cos what u get ?
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.
wait i think i see it. give me a sec
but its sin / (1-cot) you seemed to have simplified sin (1-cot)
im am to 1/sin-cos and i dont remember what is legal at this point.
ya i realized that. i would have to do the reciprocal
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?
absolutely correct!
whew! i spent a lot of time making mistakes. good practice though. its bad when your dreaming about trig identities ha
lol, but good work though :-)
thanks for the help. your a heck of a guy. i don't care what Vegeta says about you.
haha, i don't care either :P
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