how to simplify a trig identity

Question

how do you simplify this to a single trig identity:

2cos²x - cos(2x) / tan²x

Answer 1

2cos^2(x) - cos(2x)/tan^2(x)

= 2cos^2(x) - [cos^2(x) - sin^2(x)]/sin^2(x)/cos^2(x)

=2cos^2(x) - [cos^4(x) -sin^2(x)cos^2(x)]/sin^2(x)

=[2sin^2(x)cos^2(x) - cos^4(x) + sin^2(x)cos^2(x)]/sin^2(x)

=[3sin^2(x)cos^2(x) - cos^4(x)]/sin^2(x)

=3cos^2(x) - cos^4(x)/sin^2(x)

= 3cos^2(x) - cos^2(x)cos^2(x)/sin^2(x)

= 3cos^2(x) -cos^2(x)cot^2(x)

=[3-cot^2(x)]cos^2(x)

which is about as simple as it gets. It eliminates the fraction, and is in two factors with just an x angle, without the 2x.

although not that much simpler than the original 2cos^2(x) - cos(2x)/tan^2(x)

To check that this is equivalent, plug in a value for x, such as 60 degrees and see if they give the same values

cos60 = 1/2

cos120 = -1/2

cot60 = 1/sqr3

tan60 = sqr3

2cos^2(60) - cos120/tan^2(60) = 2(1/2)^2 - (-1/2)/sqr3^2 = 1/2 +(1/2)/3 = 1/2+1/6 = 2/3

(3-cot^2(x))(cos^2(x)) = (3-1/3)(1/4) = 8/3(1/4) = 2/3

It works. Odds are that's not a coincidence that they match

But the original expression, 2cos^2(x) - cos(2x)/tan^2(x) probably accidentally copied incorrectly without parentheses for the numerator. Had parentheses been put in then 2cos^2(x)-cos(2x) would be the numerator for [2cos^2(x)-cos(2x)]/tan^2(x) which = [1+cos(2x)-cos(2x)]/tan^2(x) = 1/tan^2(x) = cot^2(x)

Answer 2

2cos²x - cos(2x)/tan²x = (1 + cos(2x) - cos(2x))/tan²x = 1/tan²x = cot²x.