Trig Identities Exercise 2

Answer: 1

Solution Method: The primary Pythagorean identity \(\sin^2{x} + \cos^2{x} = 1\) comes from the application of the Pythagorean theorem on the unit circle. On the unit circle \(x=\text{cosine}\) and \(y=\text{sine},\) and the radius is 1. So \begin{align*}
    x^2+y^2 &= 1 \\[6pt]
    \cos^2{x}+\sin^2{x} &= 1
\end{align*}
The other 2 Pythagorean identities are derived from the first using the Reciprocal and Ratio Identities. 
  \begin{gather}
\sin^2{x} + \cos^2{x} = 1 \\[6pt]
1 + \tan^2{x} = \sec^2{x} \\[6pt]
1 + \cot^2{x} = \csc^2{x}
\end{gather}
We want to simplify \((\sec{x}-\tan{x})(\sec{x}+\tan{x})\) as much as possible, hopefully to a single trig function or number. First, this is a difference of squares so multiplied out it is
\begin{align*}
    (\sec{x}-\tan{x})(\sec{x}+\tan{x}) = \sec^2{x}-\tan^2{x}
\end{align*}
It's fairly clear that we're going to use the identity \(1 + \tan^2{x} = \sec^2{x}\) to simplify this further. So I substitute \(\sec^2{x}\) with \(1 + \tan^2{x}\),
\begin{align*}
    \sec^2{x}-\tan^2{x} &= 1+ \tan^2{x} -\tan^2{x} \\ 
    &= 1
\end{align*}
Therefore, I have fully simplified our original expression
\[
(\sec{x}-\tan{x})(\sec{x}+\tan{x})=1
\]
Notice I also could have rearranged the identity to match the RHS,
\begin{align*}
    1 + \tan^2{x} = \sec^2{x} \\ 
    1 = \sec^2{x} - \tan^2{x}
\end{align*}
and reached the same answer that way.