Question 405969
{{{(sin(2x)/(1+cos(2x)))*(cos(x)/(1-cos(x)))}}}
Use formulas: {{{sin (2x)=2sin(x)cos(x)}}}
___________{{{1=(sin(x))^2+(cos(x))^2}}}
___________{{{cos(2x)=(cos(x))^2-(sin(x))^2}}}
___ then {{{1+cos(2x)=(sin(x))^2+(cos(x))^2+(cos(x))^2-(sin(x))^2=2(cos(x))^2}}}
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{{{(sin(2x)/(1+cos(2x)))*(cos(x)/(1-cos(x)))=((2sin(x)cos(x))/(2(cos(x))^2))*(cos(x)/(1-cos(x)))=(sin(x)/cos(x))*(cos(x)/(1-cos(x)))=sin(x)/(1-cos(x))}}}
Multiply numerator and denominator of the fraction by {{{(1+cos(x))}}}
{{{sin(x)/(1-cos(x))=(sin(x)*(1+cos(x)))/((1-cos(x))*(1+cos(x)))=(sin(x)*(1+cos(x)))/(1-(cos(x))^2)=(sin(x)*(1+cos(x)))/(sin(x))^2=(1+cos(x))/sin(x)}}}
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we can use formulas {{{sin(x)=2sin(x/2)cos(x/2)}}}
______________{{{1-cos(x)=2(sin(x/2))^2}}}
{{{sin(x)/(1-cos(x))=(2sin(x/2)cos(x/2))/(2(sin(x/2))^2)=cos(x/2)/sin(x/2)=cot(x/2)}}}