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Prove that`(cos3x)/(sin2xsin4x)+(cos5x)/(sin4xsin6x)+(cos7x)/(sin6xsin8x)+(cos9x)/(sin8xsin10 x)`= `1/2(cos e cx)[cos e c2x-cos e c10x]`

Answer» `L.H.S. = (cos3x)/(sin2xsin4x)+(cos5x)/(sin4xsin6x)+(cos7x)/(sin6xsin8x)+(cos9x)/(sin8xsin10x)`
`=1/(2sinx)[ (2sinxcos3x)/(sin2xsin4x)+(2sinxcos5x)/(sin4xsin6x)+(2sinxcos7x)/(sin6xsin8x)+(2sinxcos9x)/(sin8xsin10x)]`
`=1/(2sinx)[ (2sinxcos3x)/(sin2xsin4x)+(2sinxcos5x)/(sin4xsin6x)+(2sinxcos7x)/(sin6xsin8x)+(2sinxcos9x)/(sin8xsin10x)]`
Using `2cosAsinB = sin(A+B) - sin(A-B)`
`=1/(2sinx)[ (sin4x-sin2x)/(sin2xsin4x)+(sin6x-sin4x)/(sin4xsin6x)+(sin8x-sin6x)/(sin6xsin8x)+(sin10x-sin8x)/(sin8xsin10x)]`
`=1/(2sinx)[ 1/(sin2x) - 1/(sin4x)+1/(sin4x) - 1/(sin6x)+1/(sin6x) - 1/(sin8x)+1/(sin8x) - 1/(sin10x)]`
`=1/(2sinx)[1/(sin2x) - 1/(sin10x)]`
`=1/2cosecx[cosec2x-cosec10x] = R.H.S.`


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