First, as in just about any trig problem, convert everything to sines and cosines:
1/cos35 1/sin55-sin35/cos35 cos55/sin551cos351sin55−sin35cos35cos55sin55
We can ignore the first part of this expression and focus on sin35/cos35 cos55/sin55sin35cos35cos55sin55, because we can see some interesting identities at play here.
Recall that cos(90-x)=sinxcos(90−x)=sinx and sin(90-x)=cosxsin(90−x)=cosx. Therefore:
sin35=cos(90-35)=cos55sin35=cos(90−35)=cos55
cos35=sin(90-35)=sin55cos35=sin(90−35)=sin55
We can now make some substitutions for sin35sin35 and cos35cos35:
sin35/cos35 cos55/sin55=cos55/sin55 cos55/sin55sin35cos35cos55sin55=cos55sin55cos55sin55
=(cos^2 55)/(sin^2 55)=cos255sin255
=cot^2 55=cot255
Now we have sec35csc55-cot^2 55sec35csc55−cot255. We can another identity on sec35csc55sec35csc55:
sec(90-x)=cscxsec(90−x)=cscx
So:
sec35=csc(90-35)=csc55sec35=csc(90−35)=csc55
Making yet another substitution, we have:
sec35csc55-cot^2 55=csc55 csc55-cot^2 55sec35csc55−cot255=csc55csc55−cot255
=csc^2 55-cot^2 55=csc255−cot255
It's looking like a Pythagorean identity might help us here...
sin^2x+cos^2x=1sin2x+cos2x=1
tan^2x+1=sec^2xtan2x+1=sec2x
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
