How do you verify `sin 2x = (2 tan x)/(1+ tan² x)`?

Modifying only the right hand side of the equation, we should first notice that the fraction's denominator is a form of the Pythagorean Identity:

`sin^2x+cos^2x=1`

`=>" "sin^2x/cos^2x+cos^2x/cos^2x=1/cos^2x`

`=>" "color(blue)(barul(|color(white)(a/a)color(black)(tan^2x+1=sec^2x)color(white)(a/a)|`

Thus, we see that we can rewrite the fraction:

`(2tanx)/(1+tan^2x)=(2tanx)/(sec^2x)`

Recalling that `" "color(green)(barul(|color(white)(a/a)color(black)(sec^2x=1/cos^2x)color(white)(a/a)|``" "`, we see that

`(2tanx)/sec^2x=(2tanx)/(1/cos^2x)`

Now, instead of dividing by `1/cos^2x`, note that this is the same as multiplying by `cos^2x/1`.

`(2tanx)/(1/cos^2x)=2tanx*cos^2x`

Rewrite `2tanx` using the identity: `" "color(purple)(barul(|color(white)(a/a)color(black)(tanx=sinx/cosx)color(white)(a/a)|`

`2tanx*cos^2x=(2sinx)/cosx*cos^2x`

The `cosx` in the denominator will cancel with one of the `cosx` terms in the numerator:

`(2sinx)/cosx*cos^2x=(2sinx)/color(red)(cancel(color(black)(cosx)))*cos^color(red)(cancel(color(black)(2)))x=2sinxcosx`

This is the identity for `sin2x`, which is the left hand side of the original equation:

`color(brown)(barul(|color(white)(a/a)color(black)(2sinxcosx=sin2x)color(white)(a/a)|`

Thus, the equation is verified.

`LHS=sin2x=(2sinxcosx)/1`

`=(2sinxcosx)/(cos^2x+sin^2x)`

`color(red)("Dividing both numerator and dinominator by "cos^2x`

`=((2sinxcosx)/cos^2x)/(cos^2x/cos^2x+sin^2x/cos^2x)`

`=(2tanx)/(1+tan^2x)=RHS`