How do you simplify #sec35csc55-tan35cot55#?

First, as in just about any trig problem, convert everything to sines and cosines:
#1/cos35 1/sin55-sin35/cos35 cos55/sin55#

We can ignore the first part of this expression and focus on #sin35/cos35 cos55/sin55#, because we can see some interesting identities at play here.

Recall that #cos(90-x)=sinx# and #sin(90-x)=cosx#. Therefore:
#sin35=cos(90-35)=cos55#
#cos35=sin(90-35)=sin55#

We can now make some substitutions for #sin35# and #cos35#:
#sin35/cos35 cos55/sin55=cos55/sin55 cos55/sin55#
#=(cos^2 55)/(sin^2 55)#
#=cot^2 55#

Now we have #sec35csc55-cot^2 55#. We can another identity on #sec35csc55#:
#sec(90-x)=cscx#

So:
#sec35=csc(90-35)=csc55#

Making yet another substitution, we have:
#sec35csc55-cot^2 55=csc55 csc55-cot^2 55#
#=csc^2 55-cot^2 55#

It's looking like a Pythagorean identity might help us here...
#sin^2x+cos^2x=1#
#tan^2x+1=sec^2x#
#underline(1+cot^2x=csc^2x)#

I've underlined that last one because it applies to our problem. To see how, simply subtract #cot^2x# in that identity:
#1=csc^2x-cot^2x#

And if this equation is supposed to hold for all values of #x#, it should hold for #x=55#. Therefore:
#sec35 csc55-tan35 cot55=csc^2 55-cot^2 55=1#