# How do you simplify sec35csc55-tan35cot55?

Jun 30, 2016

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:
$\frac{1}{\cos} 35 \frac{1}{\sin} 55 - \sin \frac{35}{\cos} 35 \cos \frac{55}{\sin} 55$

We can ignore the first part of this expression and focus on $\sin \frac{35}{\cos} 35 \cos \frac{55}{\sin} 55$, because we can see some interesting identities at play here.

Recall that $\cos \left(90 - x\right) = \sin x$ and $\sin \left(90 - x\right) = \cos x$. Therefore:
$\sin 35 = \cos \left(90 - 35\right) = \cos 55$
$\cos 35 = \sin \left(90 - 35\right) = \sin 55$

We can now make some substitutions for $\sin 35$ and $\cos 35$:
$\sin \frac{35}{\cos} 35 \cos \frac{55}{\sin} 55 = \cos \frac{55}{\sin} 55 \cos \frac{55}{\sin} 55$
$= \frac{{\cos}^{2} 55}{{\sin}^{2} 55}$
$= {\cot}^{2} 55$

Now we have $\sec 35 \csc 55 - {\cot}^{2} 55$. We can another identity on $\sec 35 \csc 55$:
$\sec \left(90 - x\right) = \csc x$

So:
$\sec 35 = \csc \left(90 - 35\right) = \csc 55$

Making yet another substitution, we have:
$\sec 35 \csc 55 - {\cot}^{2} 55 = \csc 55 \csc 55 - {\cot}^{2} 55$
$= {\csc}^{2} 55 - {\cot}^{2} 55$

It's looking like a Pythagorean identity might help us here...
${\sin}^{2} x + {\cos}^{2} x = 1$
${\tan}^{2} x + 1 = {\sec}^{2} x$
$\underline{1 + {\cot}^{2} x = {\csc}^{2} x}$

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

And if this equation is supposed to hold for all values of $x$, it should hold for $x = 55$. Therefore:
$\sec 35 \csc 55 - \tan 35 \cot 55 = {\csc}^{2} 55 - {\cot}^{2} 55 = 1$