# Thread: prove identity #3

1. ## prove identity #3

prove that (cscx-cotx)^2=(1-cosx)/(1+cosx)

my steps are as follows:
LS

(cscx-cotx)^2
(1/sinx-1/tanx)^2
(1/sinx-cosx/sinx)^2
[(1-cosx)/sinx]^2
[(1-2cosx-cos^x)/sin^2x]

now what should I do to get it to equal the other side? I am stuck here

2. Multiply by the one in the form of conjugate of the denominator over the conjugate of the denominator, i.e. 1-cos(x)/1-cos(x). In my book this gives (1-2cos(x) + cos^2(x))/(1-cos^2(x)).

Originally Posted by skeske1234
prove that (cscx-cotx)^2=(1-cosx)/(1+cosx)

my steps are as follows:
LS

(cscx-cotx)^2
(1/sinx-1/tanx)^2
(1/sinx-cosx/sinx)^2
[(1-cosx)/sinx]^2 ----------------> You are RIGHT upto here.
[(1-2cosx-cos^x)/sin^2x] <------------------------------------------> you don't need to do this step.

now what should I do to get it to equal the other side? I am stuck here
See after that,

$= \frac{(1- \cos x)^2}{\sin^2 x}$

$= \frac{(1- \cos x)^2}{1- \cos^2 x}$

$= \frac{(1- \cos x)^2}{(1- \cos x)(1 + \cos x)}$

$= \frac{1- \cos x}{1 + \cos x}$

got it ?

4. Originally Posted by skeske1234
prove that (cscx-cotx)^2=(1-cosx)/(1+cosx)

my steps are as follows:
LS

(cscx-cotx)^2
(1/sinx-1/tanx)^2
(1/sinx-cosx/sinx)^2
A[(1-cosx)/sinx]^2
[(1-2cosx-cos^x)/sin^2x]

now what should I do to get it to equal the other side? I am stuck here
I'd leave the last line off. From point A which is in red:

$\frac{(1-cos(x))^2}{sin^2(x)}$ as $sin^2(x) = 1-cos^2(x)$ we can change the denominator

$\frac{(1-cos(x))^2}{1-cos^2(x)}$

Use the difference of two squares on the bottom: $1-cos^2(x) = (1-cos(x))(1+cos(x))$

$\frac{(1-cos(x))^2}{(1-cos(x))(1+cos(x))}$.
From here you can cancel 1-cos(x) to give the final answer