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

2 Answers
May 25, 2015

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#

Jan 26, 2017

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.