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
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
I'd leave the last line off. From point A which is in red:
$\displaystyle \frac{(1-cos(x))^2}{sin^2(x)}$ as $\displaystyle sin^2(x) = 1-cos^2(x)$ we can change the denominator
$\displaystyle \frac{(1-cos(x))^2}{1-cos^2(x)}$
Use the difference of two squares on the bottom: $\displaystyle 1-cos^2(x) = (1-cos(x))(1+cos(x))$
$\displaystyle \frac{(1-cos(x))^2}{(1-cos(x))(1+cos(x))}$.
From here you can cancel 1-cos(x) to give the final answer