prove that (cos4x+cos3x+cos2x)÷(sin4x+sin3x+sin2x) = 2sinx
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Step-by-step explanation:
cos4x+cos3x+cos2x)/(sin4x+sin3x+sin2x)=cotx
LHS
=(cos4x+cos2x+cos3x)/(sin4x+sin2x+sin3x)
=[2cos3x.cosx+cos3x]/[2sin3x.cosx+sin3x]
=[cos3x(2cosx+1)]/[sin3x(2cosx+1)]
=cos3x/sin3x
=cot3x , proved
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