Prove that
`(cos 3 x)/(sin 2 xsin 4x)+(cos 5x)/(sin 4xsin 6x)+(cos 7x)/(sin 6xsin 8x)+(cos 9x)/(sin 8x sin 10x)`
`=(1)/(2) (co sec x)[co sec2x-co sec 10x]`

Explanation

Let
`f(x)=(cos 3x)/(sin 2x sin 4x)+(cos 5x)/(sin 4xsin 6x)+(cos 7x)/(sin 6xsin 8x)`
`+(cos 9x)/(sin 8x sin 10x)`
Multiply and divide by `(2sinx)` in whole expression to get
`f(x)=(sin 4x-sin2x)/(2sin x sin 2x sin 4x)+(sin6x-sin4x)/(2sin x sin 4x sin 6x)`
`+(sin 8x-sin6x)/(2sin x sin 6x sin 8x)+(sin 10x-sin 8x)/(2sin x sin 8x sin 10x)`
`=(1//2)co sec x[co sec 2x-co sec 10x]`