cos 4x + cos 3x + cos 2x / sin 4x + sin 3x + sin 2x = ?
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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.
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Answers :
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.
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