Question

How do you verify `cotx-tanx=2cot2x`?

Answer 1

Use the double angle formula for `tan(2x)` and the fact that `cot(theta) = 1/tan(theta)`

Explanation:

Double angle formula for `tan`
`color(white)("XXXX")` `tan(2x) = (2tan(x))/(1-tan^2(x))`

`2cot(2x) = 2* 1/tan(2x)`

`color(white)("XXXX")` `=2* (1-tan^2(x))/(2 tan(x))`

`color(white)("XXXX")` `=1/tan(x) - tan^2(x)/tan(x)`

`color(white)("XXXX")` `=cot(x) - tan(x)`

Answer 2

Using the relationship between tan/cot and sin-cos, plus the double angle formulae for sin and cos. (as requested)

Explanation:

Remember:

`color(red)("Basic definitions:")`
`color(white)("XXX")color(red)(tan(theta)=sin(theta)/cos(theta)color(white)("XXX")cot(theta)=cos(theta)/sin(theta))`

`color(blue)("Double angle formulae for sin and cos")`
`color(white)("XX"color(blue)(sin(2theta)=2 * sin(theta) * cos(theta)color(white)("XX")cos(2theta)=cos^2(theta)-sin^2(theta))`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Required to Prove:

`color(green)(cot(x)-tan(x)=2cot(2x)`

Proof:

`R.S.`
`color(white)("XXX")=color(green)(2cot(2x))`

`color(white)("XXX")=2 * cos(2x)/sin(2x)`

`color(white)("XXX")=(cancel2 * (cos^2(x)-sin^2(x)))/(cancel2 * sin(x) * cos(x))`

`color(white)("XXX")=cos^2(x)/(sin(x) * cos(x)) - sin^2(x)/(sin(x) * cos(x))`

`color(white)("XXX")=cos(x)/sin(x) -sin(x)/cos(x)`

`color(white)("XXX")=color(green)(cot(x)-tan(x))`

`color(white)("XXX")=L.S.`