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Exercises Trigonometry

Example 4 from Algebra and Trigonometry, 9.1 Verifying Trigonometric Identities and Using Trigonometric Identities to Simplify Trigonometric Expressions

Verifying a Trigonometric Identity Involving sec2θ

Verify the identity $\frac{\sec^{2} θ - 1}{\sec^{2} θ} = \sin^{2} θ$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
As the left side is more complicated, let’s begin there.

$\begin{matrix} \frac{\sec^{2} θ - 1}{\sec^{2} θ} & = & \frac{(\tan^{2} θ + 1) - 1}{\sec^{2} θ} & {sec}^{2} θ = \tan^{2} θ + 1 \\ = & \frac{\tan^{2} θ}{\sec^{2} θ} \\ = & \tan^{2} θ (\frac{1}{\sec^{2} θ}) \\ = & \tan^{2} θ (\cos^{2} θ) & \cos^{2} θ = \frac{1}{\sec^{2} θ} \\ = & (\frac{\sin^{2} θ}{\cos^{2} θ}) (\cos^{2} θ) & {tan}^{2} θ = \frac{\sin^{2} θ}{\cos^{2} θ} \\ = & (\frac{\sin^{2} θ}{\cos^{2} θ}) (\cos^{2} θ) \\ = & \sin^{2} θ \end{matrix}$
There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

$\begin{matrix} \frac{\sec^{2} θ - 1}{\sec^{2} θ} & = & \frac{\sec^{2} θ}{\sec^{2} θ} - \frac{1}{\sec^{2} θ} \\ = & 1 - \cos^{2} θ \\ = & \sin^{2} θ \end{matrix}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Show that $\frac{\cot θ}{\csc θ} = \cos θ .$

How did it go?

PreviousVerifying the Equivalency Using the Even-Odd Identities Next Creating and Verifying an Identity

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 9.1, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.