Question

cos4 x + cos3 x + cos2 x/
sin4 x + sin3 x + sin2 x = cot3 x​

Answer 1

Answer:

R.H.S

Step-by-step explanation:

L.H.S

=cos4x+cos3x+cos2x/sin4x+sin3x+sin2x

=(cos4x+cos2x)+cos3x/(sin4x+sin3x)+sin3x

=2cos[4x+2x/2]cos[4x-2x/2]+cos3x/2sin[4x+2x/2]cos[4x-2x/2]+sin3x

=[cosA+cosB=2cos[A+B/2]cos[A-B/2]sinA+sinB=2sin[A+B/2]cos[A-B/2]]

=2cos3xcosx+cos3x/2sin3xcosx+sin3x

=cos3x(2cosx+1)/sin3x(2cosx+1)

=cot3x

=R.H.S