Question

Hi this is an equation from my pre-calc class and i don't really understand the steps in the conversion and factoring parts.


`cot^(3) x+cot^(2) x+cotx+1`

Answer 1

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)

Answer 2

Given:
`cot^(3) x+cot^(2) x+cotx+1`

Factor:

`cot {}^(2) (x)(cot(x) + 1) + 1(cot(x) + 1`

`(cot {}^(2) (x) + 1)(cot(x) + 1)`

Substitute
`\cot {}^(2) (x) + 1 = csc {}^(2) (x):csc {}^(2) (x)(cot(x) + 1)`