cos10xcos2x sin2xsin10xSolutioncosA cosB 2 cosA B2 cosA

cos(10x)+cos(2x) / sin(2x)-sin(10x)

Solution

cosA + cosB = 2 cos[(A + B)/2] cos[(A - B)/2]

sinA - sinB = 2 sin[(A - B)/2] cos[(A + B)/2]

cos(10x) + cos(2x) = 2 cos[(10x + 2x)/2] cos[(10x - 2x)/2]

==> cos(10x) + cos(2x) = 2 cos[(12x)/2] cos[(8x)/2] = 2 cos(6x) cos(4x)

sin(2x) - sin(10x) = 2 sin[(2x - 10x)/2] cos[(2x + 10x)/2]

==> sin(2x) - sin(10x) = 2 sin[-8x/2] cos[12x/2]

==> sin(2x) - sin(10x) = -2 sin(4x) cos(6x) ; since sin(-x) = -sinx

[ cos(10x) + cos(2x) ] / [ sin(2x) - sin(10x) ] = 2 cos(6x) cos(4x) / [ -2 sin(4x) cos(6x) ]

==> - cos(4x) / sin(4x)

==> -cot(4x)

Hence [ cos(10x) + cos(2x) ] / [ sin(2x) - sin(10x) ] = -cot(4x)