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#