Question 644034
{{{(1-cos2x)/tanx = sin2x}}}

start with left side and prove it's equal to right side


{{{(1-cos2x)/tanx }}} ....use identity {{{1/tan(x)=ctg(x)}}}


={{{(1-cos2x)ctg(x)}}}....use identity {{{cot(x)=-(sin(2x)/(cos(2x)-1))}}}


={{{(1-cos2x)*(-(sin(2x)/(cos(2x)-1)))}}}....pull {{{-1}}} in front of {{{(1-cos2x)}}}


={{{(1-cos2x)*(-(sin(2x)/(-(1-cos(2x)))))}}}


={{{(1-cos2x)*(sin(2x)/(1-cos(2x))))}}}


={{{cross((1-cos2x))*(sin(2x)/cross((1-cos(2x)))))}}}


={{{sin(2x)}}}.....we arrived to the right side of your identity and it proves your identity