Question

How do I prove:
sinB / 1 + cosB + 1+cosB / sinB = 2cscB

Answer 1

\[\frac{ sinB }{ 1 + cosB } + \frac{ 1+cosB }{ sinB } = 2 cscB\]

Answer 2

did you try finding a common denominator?

Answer 3

well...i see imma work on the left side and the common denominator would be \[1+cosB\] right?

Answer 4

I think it would be (1+cosB) sinB

Answer 5

multiply the first fraction by sinB/sinB
and the 2nd fraction by (1+cosB)/(1+cosB)

Answer 6

if thats so, at least I got the 1 + cosB right

Answer 7

okay @phi Let me write that down really fast and ill post it on here in a sec

Answer 8

okay @phi \[\frac{ \sin ^{2}B }{ sinB+sinBcosB } + \frac{ \cos ^{2}B +2cosB+1 }{ sinB+sinBcosB }\]

Answer 9

and i can combine them to be....

Answer 10

yes, but I would not have bothered to multiply the bottoms... because chances are we are going to cancel something...

Answer 11

@lilsis76 ... don't multiply sinB with 1+cosB. Keep it as
\[sinB(1+cosB)\] easier like that

Answer 12

and i can combine them to be....\[\frac{ \sin^2 B + \cos^2 B + 2 \cos B + 1 }{ sinB + sinBcosB }\]

Answer 13

yes, now use c^2 + s^2 = 1

Answer 14

okay I have this then @phi

Answer 15

\[\frac{ 1 + 2cosB + 1 }{ sinB + sinBcosB }\]

Answer 16

i get \[\frac{ 2cosB +2 }{ \sin B + sinBcosB }\] right? @phi

Answer 17

yes, now factor out a 2 up top
and undo your work below (factor out the sin)

Answer 18

the 1 + cos B can = sinB so i can get....@phi