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

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SUMMARY

The expression $\frac{\sec\theta}{\tan\theta}$ simplifies directly to $\csc(\theta)$ by converting secant and tangent into sine and cosine functions. Specifically, $\frac{\sec(\theta)}{\tan(\theta)}$ is rewritten as $\frac{\frac{1}{\cos(\theta)}}{\frac{\sin(\theta)}{\cos(\theta)}}$, which simplifies to $\frac{1}{\sin(\theta)}$, yielding $\csc(\theta)$. This simplification is valid under the conditions that $\sin(\theta) \neq 0$ and $\cos(\theta) \neq 0.

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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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