Question

prove that, cos4x+cos3x+cos2x/sin4x+sin3x+sin2x=cos2x​

Answer 1

Answer requested are as follows---

(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

Answer 2

Answer:

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