LaddiusMaximus
  • LaddiusMaximus
(sinx/(1+cosx)) +((1+cosx)/sinx) totally lost
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
What do you have to do? prove it?
LaddiusMaximus
  • LaddiusMaximus
yeah=2cscx
myininaya
  • myininaya
combine the fractions by finding a common denominator

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More answers

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

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