Question

How do you simplify `sec35csc55-tan35cot55`?

Answer 1

Apply a bunch of complementary angle identities to get a result of `1`.

Explanation:

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`