MHB How Can $\frac{\sec\theta}{\tan\theta}$ Be Simplified to $\csc(\theta)$?

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The expression $\frac{\sec\theta}{\tan\theta}$ can be simplified to $\csc(\theta)$ by converting secant and tangent into sine and cosine functions. This involves rewriting $\sec(\theta)$ as $\frac{1}{\cos(\theta)}$ and $\tan(\theta)$ as $\frac{\sin(\theta)}{\cos(\theta)}$. By performing the division, the equation simplifies to $\frac{1}{\sin(\theta)}$, which is equal to $\csc(\theta)$. It's important to note that this simplification is valid under the conditions that $\sin(\theta) \neq 0$ and $\cos(\theta) \neq 0$. Thus, $\frac{\sec\theta}{\tan\theta} = \csc(\theta)$ when these restrictions are met.
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I have $\frac{sec\theta}{tan\theta}$. How can I simplify it to get $\csc\left({\theta}\right)$ ?
 
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tmt said:
I have $\frac{sec\theta}{tan\theta}$. How can I simplify it to get $\csc\left({\theta}\right)$ ?
I prefer to change everything to sines and cosines.
[math]\frac{sec( \theta )}{tan( \theta )} = \frac{ \frac{1}{cos( \theta )}}{ \frac{sin( \theta )}{cos( \theta )}}[/math]

[math]= \frac{ \frac{1}{cos( \theta )}}{ \frac{sin( \theta )}{cos( \theta )}} \cdot \frac{ cos( \theta )}{cos( \theta )}[/math]

[math]= \frac{1}{sin( \theta )} = csc( \theta )[/math]

With, of course, the restrictions that [math]sin( \theta ) \neq 0[/math] and [math]cos( \theta ) \neq 0[/math]

-Dan
 

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