pancha3

2021-08-21

Find $\frac{1}{\mathrm{sec}\theta -\mathrm{tan}\theta }+\frac{1}{\mathrm{sec}\theta +\mathrm{tan}\theta }$

Nicole Conner

We are given: $\frac{1}{\mathrm{sec}\theta -\mathrm{tan}\theta }+\frac{1}{\mathrm{sec}\theta +\mathrm{tan}\theta }$

Rewrite using the LCD which is $\left(\mathrm{sec}\theta -\mathrm{tan}\theta \right)\left(\mathrm{sec}\theta +\mathrm{tan}\theta \right)=\frac{1}{\mathrm{sec}\theta -\mathrm{tan}\theta }\left(\frac{\mathrm{sec}\theta +\mathrm{tan}\theta }{\mathrm{sec}\theta +\mathrm{tan}\theta }\right)+\frac{1}{\mathrm{sec}\theta +\mathrm{tan}\theta }\left(\frac{\mathrm{sec}\theta -\mathrm{tan}\theta }{\mathrm{sec}\theta -\mathrm{tan}\theta }\right)=\frac{\left(\mathrm{sec}\theta +\mathrm{tan}\theta \right)+\left(\mathrm{sec}\theta -\mathrm{tan}\theta \right)}{{\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta }$

Recall that ${\mathrm{sec}}^{2}\theta =1+{\mathrm{tan}}^{2}\theta$ so ${\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta =1=2\frac{\mathrm{sec}}{\theta }1=2\mathrm{sec}\theta$

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