Question 131980
SOLUTION -
 
 {{{(Sin2x+sin4x)/(cos2x-cos4x)}}}=cotx
 Taking L.H.S we have
 [{{{2sin((2x+4x)/2)}}}{{{cos((2x-4x)/2)}}}]/[{{{-2sin((2x+4x)/2)}}}{{{sin((2x-4x)/2)}}}]    
   { (since {{{sinc+sind}}}=2{{{sin((c+d)/2)}}}{{{cos((c-d)/2)}}}
           {{{cosc-cosd}}}=-2{{{sin((c+d)/2)}}}{{{sin((c-d)/2)}}}}
 ={{{(2sin3xcos(-x))}}}/{{{(-2sin3xsin(-x))}}}
  ={{{cosx/sinx}}}    {since cos(-x)=cosx and sin(-x)=-sinx}
 = cotx
 
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