If you're just factorizing, you can do so by grouping.
cot³(x) + cot²(x) + cot(x) + 1
= cot²(x) (cot(x) + 1) + cot(x) + 1
= (cot²(x) + 1) (cot(x) + 1)
Put another way, if y = cot(x) + 1, then
cot²(x) y + y = (cot²(x) + 1) y
We can simplify this somewhat. Recall the Pythagorean identity,
sin²(x) + cos²(x) = 1
Dividing through both sides of the equation by sin²(x) reveals another form of the identity,
sin²(x)/sin²(x) + cos²(x)/sin²(x) = 1/sin²(x)
1 + cot²(x) = csc²(x)
Then we end up with
cot³(x) + cot²(x) + cot(x) + 1 = csc²(x) (cot(x) + 1)
Given:[tex]cot^{3} x+cot^{2} x+cotx+1[/tex]
Factor:
[tex]cot {}^{2} (x)(cot(x) + 1) + 1(cot(x) + 1[/tex]
[tex](cot {}^{2} (x) + 1)(cot(x) + 1)[/tex]
Substitute [tex] \cot {}^{2} (x) + 1 = csc {}^{2} (x):csc {}^{2} (x)(cot(x) + 1)[/tex]
