Question

how to prove the identity of 1-cos(x)/sin(x) + sin(x)/1+cos(x) = 2 csc (x)

Answer 1

\[\frac{1-cos(x)}{sin(x)}+\frac{sin(x)}{1+cos(x)}=2csc(x)\]

Answer 2

?

Answer 3

@cgurley86

Answer 4

yeah thats the equation

Answer 5

its not

Answer 6

are you sure its 1+cos(x) in the second term?

Answer 7

yeah thats what is written on my hw

Answer 8

the left hand side at x=pi/3 is 2/sqrt{3}
the right hand side at x = pi/3 is 4/sqrt(3)

Answer 9

im confused?

Answer 10

im saying if you let x =pi/3 on the left side of the equation we get a different number than doing the same thing on the right side of the equation

Answer 11

\[\frac{1-cos(x)}{sin(x)}+\frac{sin(x)}{1-cos(x)}=2csc(x)\]
this is true

Answer 12

must have been a typo

Answer 13

\[\frac{1-cos(x)}{1-cos(x)}*\frac{1-cos(x)}{sin(x)}+\frac{sin(x)}{sin(x)}*\frac{sin(x)}{1-cos(x)}\]
=
\[\frac{(1-cos(x))^2+sin^2(x)}{(1-cos(x))*sin(x)}=\frac{1-2cos(x)+[pcos^2(x)+sin^2(x)]}{(1-cos(x))*sin(x)}\]=\[\frac{1-2cos(x)+[1]}{(1-cos(x))*sin(x)}=\frac{2-2cos(x)}{(1-cos(x))*sin(x)}=\frac{2(1-cos(x))}{(1-cos(x))*sin(x)}=\frac{2}{sin(x)}=\\2csc(x)\]

Answer 14

it should say this where it gets cut off
\[\frac{2}{sin(x)}=2csc(x)\]

Answer 15

let me know if you don't understand anything

Answer 16

that pcos(x) should just be cos(x)

Answer 17

understand @cgurley86

Answer 18

not really

Answer 19

which part?

Answer 20

notice I just multiplied each term by 1?

Answer 21

no