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Sir / Madam ,
Prove that : – cos 4x + cos 3x+ cos 2x / sin 4x + sin 3x + sin 2x = cot 3x ..

Jeeva , 9 Years ago
Grade 11
anser 2 Answers
Himanshu

Last Activity: 9 Years ago

L.H.S. = (cos4x + cos2x) +cos3x / (sin4x + sin2x) + sin3x
          = 2cos[(4x+2x)/2].cos[(4x-2x)/2] + cos3x / 2sin[(4x+2x)/2].cos[(4x-2x)/2] + sin3x
          = 2cos3x.cosx + cos3x / 2sin3x.cosx + sin3x
          = cos3x(2cosx + 1) / sin3x(2cosx + 1)                   // Taking common
          = cos3x(2cosx + 1) / sin3x(2cosx + 1)
          = cos3x / sin3x
          = cot3x
Hence, L.H.S = R.H.S.

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the answer to your question.
 
L.H.S. = (cos4x + cos2x) +cos3x / (sin4x + sin2x) + sin3x
          = 2cos[(4x+2x)/2].cos[(4x-2x)/2] + cos3x / 2sin[(4x+2x)/2].cos[(4x-2x)/2] + sin3x
          = 2cos3x.cosx + cos3x / 2sin3x.cosx + sin3x
          = cos3x(2cosx + 1) / sin3x(2cosx + 1)                  
          = cos3x(2cosx + 1) / sin3x(2cosx + 1)
          = cos3x / sin3x
          = cot3x
Hence, L.H.S = R.H.S.
 
Thanks and regards,
Kushagra

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