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

1 Answer

It is

Write #csc^4 x# as #1/(sin^4 x)# and #cot^4 x# as #(cos^4 x)/(sin^4 x)#
#=> csc^4x - cot^4x = 1/(sin^4 x) - (cos^4 x)/(sin^4 x) = (1 - cos^4 x)/(sin^4 x)#

Now recall that #a^2 - b^2 = (a - b)*(a + b)# and use this fact with #a^2# being and
#b^2# being #cos^4 x# so that #a# is 1 and b is #cos^2 x#

So
#csc^4x - cot^4x = (1 - cos^4 x)/(sin^4 x) = (1 - cos^2 x)*(1 + cos^2 x) / sin^4 x#

But #cos^2 x + sin^2 x = 1# so that #1 - cos^2 x = sin^2 x#
so #csc^4x - cot^4x = sin^2 x * (1 + cos^2 x) / sin^4 x = (1 + cos^2 x) / sin^2 x = (1/sin^2 x) + (cos^2 x / sin^2 x) = csx^2 x + cos^2 x #