Question

How do you prove `csc^4x-cot^4x=csc^2x+cot^2x`?

Answer 1

Left side:

`1/sin^4 x - cos^4 x/sin^4 x = (1 - cos^4 x)/(sin^4 x)`=

`= [(1- cos^2 x)(1 + cos^2 x)]/(sin^4 x) = ((sin^2 x)(1 + cos ^2 x))/(sin^4 x)`
`= (1 + cos^2x)/sin^2 x = 1/(sin^2 x)+ (cos^2 x)/(sin^2 x) =`
`= csc^2 x + cot^2 x`

Answer 2

Using the Identity `; csc^2y-cot^2y=1`,

The L.H.S. `=csc^4x-cot^4x=(csc^2x-cot^2x)(csc^2x+cot^2x)`
`=csc^2x+cot^2x=`the R.H.S.