precalculus: factoring

Question

Factor and simplify.  (cos^2θ-1)/(cos^2θ-cosθ)

Answer 1

Let    x=θ

 

(cos²x - 1) / (cos²x - cosx)

 

 

Multiply by the conjugate of the denominator.

 

(cos²x - 1)(cos²x + cosx)

________________________ =

 (cos²x - cosx)(cos²x + cosx)

 

 

 

    cos⁴x + cos³x - cos²x - cosx

_____________________________  =

       cos⁴x - cos²x

 

 

Factor by top and bottom.

 

cosx(cos³x + cos²x - cosx - 1)

_________________________  =

(cos²x - cosx)(cos²x + cosx)

 

 

Factor by grouping on the top.

 

cosx [cos²(cosx + 1) - (cosx + 1)]

____________________________ =

   (cos²x - cosx)(cos²x + cosx)

 

 

 

cosx(cos²x - 1)(cosx + 1)

____________________________ =

cosx(cosx - 1)* cosx(cosx + 1)

 

 

 

cosx(cosx + 1)(cosx - 1)(cosx + 1)

______________________________  =

cos²x(cosx - 1)(cosx + 1)

 

 

 

(cosx + 1) / cosx  --->  final solution.

 

Answer 2

factor both terms and simplify

(cos-1)(cos+1)/(cos)(cos-1)

simplify(cancel common factors)

(cosθ+1)/cosθ

or...

cos/cos + 1/cos

1+secθ (either answer is correct)

Answer 3

The numerator is a difference of squares.  So, using the format

a² - b² = (a - b)(a + b), we have

(cosΘ - 1)(cosΘ + 1).

 

For the denominator, factor a cosΘ out, yielding

(cosΘ)(cosΘ - 1)

 

After cancelling (cosΘ - 1), we are left with

(cosΘ + 1) / (cosΘ).

 

Hope that helps.  Thank you for the question!