Question

(sinx/(1+cosx)) +((1+cosx)/sinx) totally lost

Answer 1

What do you have to do? prove it?

Answer 2

yeah=2cscx

Answer 3

combine the fractions by finding a common denominator

Answer 4

yeah i got that, but what do I use, sinx? or 1+cosx?

Answer 5

I'm working on it

Answer 6

The Common Denominator will be sinx(1+cosx)

Answer 7

what do you mean? how to combine these fractions: 2/3+5/7

Answer 8

so the common denominator is (sinx)1+cosx?

Answer 9

sinx(1+cosx)

Answer 10

yes that is the common denom

Answer 11

okay so now I have (sinx^2x(1+cosx)/sinx(1+cosx)) + (sinx(2+cos^2x)/sinx(1+cosx))?

Answer 12

you will have (sinx)^2+(1+cosx)^2 in the numerator and sinx(1+cosx) is the denominator

Answer 13

sin^2x on both sides?

Answer 14

I think you should have this so far: \[\sin^2x + (1+cosx)^2/sinx(1+cosx)\]

Answer 15

ok look at your first fraction. what does the bottom lack so that it has the same denominator as the 2nd. sinx, right?

Answer 16

multiply both the top and bottom by sinx on the first fraction

Answer 17

hmm i had [\sin^2x(1+cosx)+sinx(2+\cos^2x)/sinx(1+cosx)\]

Answer 18

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?

Answer 19

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

Answer 20

I follow you so far.

Answer 21

so we have [sinx*sinx+(1+cosx)(1+cosx)]/[sinx(1+cosx)]

Answer 22

[(sinx)^2+1+2cosx+(cosx)^2]/[sinx(1+cosx)]

Answer 23

I think I got the complete answer/proof

Answer 24

what is (sinx)^2+(cosx)^2

Answer 25

=1

Answer 26

lol gj, but i was laddius

Answer 27

oops sorry

Answer 28

but i was asking*

Answer 29

do you still follow laddius

Answer 30

so the numerator becomes sin^2x + (1+cosx)^2?

Answer 31

yes

Answer 32

now foil that (1+cosx)^2

Answer 33

exactly lol

Answer 34

so now its 1+2cosx+cosx^2?

Answer 35

yes!

Answer 36

+(sinx)^2

Answer 37

now do you see (sinx)^2+(cosx)^2 anywhere in that numerator

Answer 38

ahhhhh pythagorean

Answer 39

what do you have now?

Answer 40

so now I have 2+2cosx/sinx(1+cosx)

Answer 41

ok what does the numerator terms have in common

Answer 42

2?

Answer 43

HAHA ignore my last 3 posts..

Answer 44

does anything cancel?

Answer 45

jp?

Answer 46

the 2s cancel, right?

Answer 47

no the 1+cosx do

Answer 48

so you have 2/sinx

Answer 49

which is ....

Answer 50

I have 2+2cosx in the numerator

Answer 51

cosx+cosx= 2cosx, right?

Answer 52

didnt you say the numerator terms had a 2 in common so factor that out in you have 2(1+cosx)/[sinx(1+cosx)]

Answer 53

crap then the (1+cosx) cancel out and im left with 2cscx

Answer 54

BINGO!