# Where defined, \frac{sin^2(\theta)}{cos^2(\theta)} = 1?

## Question:

Where defined, {eq}\frac{sin^2(\theta)}{cos^2(\theta)} = 1? {/eq}

## Rational Function:

A rational function is of the form {eq}\displaystyle \frac{p}{q} {/eq}, where p and q are defined functions and {eq}q \neq 0 {/eq}

Also, note that {eq}\cos \theta = 0 \Rightarrow \theta = \neq 2n\pi + \frac{\pi}{2} ;\ 2n\pi + \frac{3\pi}{2} {/eq}

## Answer and Explanation:

We are given that {eq}\displaystyle \frac{\sin^2(\theta)}{\cos^2(\theta)} = 1 {/eq}

As, we know that any rational function {eq}\displaystyle \frac{p}{q} {/eq} is defined if {eq}q \neq 0 {/eq}

So, the given function {eq}\displaystyle \frac{\sin^2(\theta)}{\cos^2(\theta)} = 1 {/eq} is defined only if

{eq}\displaystyle \cos^2(\theta) \neq 0 \\ \Rightarrow \displaystyle \theta \neq 2n\pi + \frac{\pi}{2} ;\ 2n\pi + \frac{3\pi}{2} {/eq}

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from GMAT Prep: Help and Review

Chapter 10 / Lesson 11