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##