Question

(1-3cosx-4cos^2x/sin^2x=1-4cosx/1-cosx

Answer 1

do what? prove? solve?

Answer 2

prove

Answer 3

Hint first, ok? denominator : sin^2 = 1- cos^2 = (1-cos) \(\color{red}{(1+cos)}\)
pay attention at the red part, we are gonna cancel it out with numerator

Answer 4

numerator: \(1 - 3cos-4cos^2 = cos^2 \color {blue}{+2cos}\) +1\(\color{blue}{-2cos}-3cos-5cos^2\)
know why I do it? for the blue part, I add, then subtract, so nothing happen, right?
for the black part, I just do -4cos^2 = cos^2 - 5cos^2, and put cos^2 at front and -5cos^2 at last. got me?

Answer 5

so far, the first 3 terms I put forms the form of (cos +1)^2
the last 3 terms becomes -5cos -5cos^2. I factor those terms by -5cos(cos +1)

Answer 6

now combine!!!
\((cos +1)^2-5cos(cos +1)\)
= \(\color{red}{(cos +1)}(cos +1 -5cos)\)
=\(\color{red}{(cos+1)}(1-4cos)\)
that's the numerator

Answer 7

combine the nume and the denominator , cancel the red part, you get the right hand side
the proof done

Answer 8

The left side can be factored as follows:
Numerator: -4cos^2 x - 3cos x + 1 = (cos x + 1)(1 - 4cos x)
Denominator: sin^2 x = 1 - cos^2 x = (1 - cos x)(1 + cox )
After simplification get the answer.