# Simplify (\sec^2\theta - 1)(\csc^2\theta\cos^2\theta).

## Question:

Simplify

{eq}(\sec^2\theta - 1)(\csc^2\theta\cos^2\theta). {/eq}

## Simplifying Trigonometric Expressions:

In trigonometry, a trigonometric identity is a mathematical equation setting two trigonometric expressions equal to one another. We can use trigonometric identities, along with some algebra, to simplify trigonometric expressions.

## Answer and Explanation:

The expression {eq}\left ( sec^{2}\theta -1 \right )\left ( csc^{2}\theta \: cos^{2}\theta \right ) {/eq} simplifies to 1. We can simplify this expression using the following trigonometric identities.

- {eq}\displaystyle csc\theta =\frac{1}{sin\theta } {/eq}

- {eq}\displaystyle sec\theta =\frac{1}{cos\theta } {/eq}

- {eq}1-cos^{2}\theta =sin^{2}\theta {/eq}

We start with the given expression.

- {eq}\displaystyle \left ( sec^{2}\theta -1 \right )\left ( csc^{2}\theta \: cos^{2}\theta \right ) {/eq}

Now, we use the first two trigonometric identities to rewrite {eq}csc^{2}\theta {/eq} and {eq}sec^{2}\theta {/eq}, and simplify.

- {eq}\displaystyle \left ( \frac{1}{cos^{2}\theta}-1 \right )\left ( \frac{1}{sin^{2}\theta }\cdot cos^{2}\theta \right )=\left ( \frac{1}{cos^{2}\theta}-1 \right )\left ( \frac{cos^{2}\theta }{sin^{2}\theta } \right ) {/eq}

We can now multiply {eq}\displaystyle \frac{cos^{2}\theta }{sin^{2}\theta } {/eq} by {eq}\displaystyle \left ( \frac{1}{cos^{2}\theta}-1 \right ) {/eq} to get the following:

- {eq}\displaystyle \left ( \frac{1}{cos^{2}\theta}-1 \right )\left ( \frac{cos^{2}\theta }{sin^{2}\theta } \right )=\frac{1}{sin^{2}\theta }-\frac{cos^{2}\theta }{sin^{2}\theta } {/eq}

We will now subtract the two fractions. Since the denominators are the same, we simply subtract their numerators and leave the denominator as is to get the following:

- {eq}\displaystyle \frac{1}{sin^{2}\theta }-\frac{cos^{2}\theta }{sin^{2}\theta }=\frac{1-cos^{2}\theta }{sin^{2}\theta } {/eq}

We can now use the third trigonometric identity to rewrite {eq}1-cos^{2}\theta {/eq} as {eq}sin^{2}\theta {/eq}, and then simplify.

- {eq}\displaystyle \frac{1-cos^{2}\theta }{sin^{2}\theta }=\frac{sin^{2}\theta }{sin^{2}\theta }=1 {/eq}

All together, we get that {eq}\left ( sec^{2}\theta -1 \right )\left ( csc^{2}\theta \: cos^{2}\theta \right )=1 {/eq}.

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from Honors Precalculus Textbook

Chapter 23 / Lesson 1