How do you verify (tanx/(1+cosx)) + (sinx/(1-cosx))=cotx + secxcscx?

TL;WR – Just find the LCD of the two fractions on the left side and proceed to simplify.
What I like to do when I have two dissimilar fractions (ones with different denominators) is this:
##(tan x/(1 + cos x)) * (1 – cos x)/(1- cos x) + (sin x/(1 – cos x)) * (1 +cos x)/(1+ cos x) = cot x + secxcscx##
Remember that ##(1 – cos x)/(1- cos x)## and ##(1 +cos x)/(1+ cos x)## are equal to one so multiplying anything by them won’t change the value thus preserving the equals sign.
So if we do that…
##((tan x* (1 – cos x))/((1 + cos x) * (1- cos x))) + ((sin x* (1 +cos x))/((1 – cos x) * (1+ cos x))) = cot x + secxcscx##
Now that’s a mouthful. Let’s simplify some stuff first.
##tan x * (1 – cos x) = tan x – tan x cos x = tanx – sinx##
##sin x* (1 +cos x) = sin x + sin x cos x##
##(1 – cos x) * (1+ cos x) = 1^2 – cos^2 x = 1 – cos ^2x = sin^2 x##
(by the difference of squares pattern but it won’t always be like this you could get a trinomial for example)
So let’s substitute those answers into our equation:
##(tanx – sinx)/(sin^2 x) + (sin x + sin x cos x)/(sin^2 x) = cot x + secxcscx##
There shorter and simpler. We still have a bit of a ways to go but props for sticking it out so far. So then we combine both of our fractions on the left side because now they have the same denominator.
##(tanx – cancel(sinx) + cancel(sin x) + sin x cos x)/(sin^2 x) = cot x + secxcscx##
We have two ##sin## but of different signs (get it?):
##(tanx + sin x cos x)/(sin^2 x) = cot x + secxcscx##
Separate the left side into two terms:
##tanx/sin^2x + (sinxcosx)/sin^2x = cot x + secxcscx##
Then let’s have some simplification:
##tanx/sin^2x = (sinx/cosx) / sin^2x = cancel(sinx)/cosx * 1/cancel(sin)^2x = 1/(cosxsinx) = secxcscx##
##(cancel(sinx)cosx)/cancel(sin)^2x = cosx/sinx = cotx##
So if we substitute:
##secxcscx + cotx = cotx + secxcscx##