How do I prove this? cot(x)(1-cos(2x))=sin(2x)

`LHS=cotx(1-cos2x)`

`=cosx/sinx*2sin^2x`

`=2sinx*cosx=sin2x=RHS`

`color(green)(N.B: cos(2x) = cos^2x - sin^2x`

`color(green)(sin(2x) = 2sinxcosx`

`cot(x) = 1/tan(x) = 1/ (sinx/cosx) = cos(x)/sin(x)`

`cot(x)(1-cos(2x))`

`=> [cos(x)/sin(x)] [1- ( cos^2x - sin^2x`)]

`=> [cos(x)/sin(x)] [1- cos^2x + sin^2x`]

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

`=> [cos(x)/sin(x)] [2sin^2x`]

`=>2sinxcosx`

Since

`sin(2x) = 2sinxcosx`

Hence,

`color(crimson)(cot(x)(1-cos(2x)) = sin(2x)`

`Q. E. D`

convert `cotx` into sins and cosines with the identity

`cotx=cosx/sinx`

`cosx/sinx(1-cos2x)=sin2x`

turn `sin2x` in terms of a single multiple of `x` using the double angle formula

`sin2x=2cosxsinx`

`cosx/sinx(1-cos2x)=2cosxsinx`

expand the brackets

`cosx/sinx+(-cosx*cos2x)/sinx=2cosxsinx`

using one of the double angle formula for cosine

`cos2x=1-2sinx`

substitute

`cosx/sinx+(-cosx(1-2sin^2x))/sinx=2cosxsinx`

expand the brackets

`cosx/sinx+(-cosx+2cosxsin^2x)/sinx=2cosxsinx`

add the fractions

`(cosx-cosx+2cosxsin^2x)/sinx=2cosxsinx`

cancel `cosx`

`(cancel(cosx-cosx)+2cosxsin^2x)/sinx=2cosxsinx`

`(2cosxsin^cancel(2)x)/cancelsinx=2cosxsinx`

`2cosxsinx=2cosxsinx`

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)cotx=cosx/sinx`

`•color(white)(x)cos2x=2cos^2x-1" and "sin2x=2sinxcosx`

`•color(white)(x)sin^2x+cos^2x=1`

`"consider the left side"`

`rArrcosx/sinx(1-(2cos^2x-1))`

`=cosx/sinx(2-2cos^2x)`

`=cosx/sinx(2(1-cos^2x))`

`=cosx/sinx(2sin^2x)`

`=2sinxcosx`

`=sin2x="right side "rArr"verified"`