Scroll to Top

Virtual Math Learning Center

Virtual Math Learning Center Texas A&M University Virtual Math Learning Center
tags:
Fundamental Trigonometric Identity Trigonometry Trigonometry Series

Trig Identities Exercise 2

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Use the Pythagorean Identities to simplify
\[
(\sec{x}-\tan{x})(\sec{x}+\tan{x})
\]

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. 
 
Pythagorean 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.

See more videos from this section

Prev Video Return to Section