1.

(cos(3*x) %2B cos(4*x) %2B cos(2*x))/(sin(3*x) %2B sin(4*x) %2B sin(2*x))=cot(3*x)

Answer»

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.


Discussion

No Comment Found

Related InterviewSolutions