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What do you have to do? prove it?
combine the fractions by finding a common denominator
yeah i got that, but what do I use, sinx? or 1+cosx?
I'm working on it
The Common Denominator will be sinx(1+cosx)
what do you mean? how to combine these fractions: 2/3+5/7
so the common denominator is (sinx)1+cosx?
yes that is the common denom
okay so now I have (sinx^2x(1+cosx)/sinx(1+cosx)) + (sinx(2+cos^2x)/sinx(1+cosx))?
you will have (sinx)^2+(1+cosx)^2 in the numerator and sinx(1+cosx) is the denominator
sin^2x on both sides?
I think you should have this so far: \[\sin^2x + (1+cosx)^2/sinx(1+cosx)\]
ok look at your first fraction. what does the bottom lack so that it has the same denominator as the 2nd. sinx, right?
multiply both the top and bottom by sinx on the first fraction
hmm i had [\sin^2x(1+cosx)+sinx(2+\cos^2x)/sinx(1+cosx)\]
now look at the 2nd fraction what does the bottom lack so that it has the same denominator as the first fraction, (1+cosx), right?
so multiply the top and bottom of the 2nd fraction by (1+cosx). Now the fractions have the same denominator, you can write it as one fraction now
I follow you so far.
so we have [sinx*sinx+(1+cosx)(1+cosx)]/[sinx(1+cosx)]
I think I got the complete answer/proof
what is (sinx)^2+(cosx)^2
lol gj, but i was laddius
but i was asking*
do you still follow laddius
so the numerator becomes sin^2x + (1+cosx)^2?
now foil that (1+cosx)^2
so now its 1+2cosx+cosx^2?
now do you see (sinx)^2+(cosx)^2 anywhere in that numerator
what do you have now?
so now I have 2+2cosx/sinx(1+cosx)
ok what does the numerator terms have in common
HAHA ignore my last 3 posts..
does anything cancel?
the 2s cancel, right?
no the 1+cosx do
so you have 2/sinx
which is ....
I have 2+2cosx in the numerator
cosx+cosx= 2cosx, right?
didnt you say the numerator terms had a 2 in common so factor that out in you have 2(1+cosx)/[sinx(1+cosx)]
crap then the (1+cosx) cancel out and im left with 2cscx