acidizihvzs

2022-04-04

Evaluate:
$\frac{\mathrm{cos}\left(90+\theta \right)+\mathrm{cot}\left(450+\theta \right)}{\mathrm{csc}\left(450-\theta \right)-\mathrm{tan}\left(180+\theta \right)}+\frac{\mathrm{tan}\left(180+\theta \right)+\mathrm{sec}\left(180-\theta \right)}{\mathrm{tan}\left(360-\theta \right)-\mathrm{sec}\left(-\theta \right)}$

$1+\frac{\mathrm{sec}\theta -\mathrm{tan}\theta }{\mathrm{sec}\theta +\mathrm{tan}\theta }$
$=1+\frac{\mathrm{sec}\theta -\mathrm{tan}\theta }{\mathrm{sec}\theta +\mathrm{tan}\theta }\cdot \frac{\mathrm{cos}\theta }{\mathrm{cos}\theta }$
$=1+\frac{\frac{1}{\mathrm{cos}\theta }-\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }}{\frac{1}{\mathrm{cos}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }}\cdot \frac{\mathrm{cos}\theta }{\mathrm{cos}\theta }$
$=1+\frac{1-\mathrm{sin}\theta }{1+\mathrm{sin}\theta }$
$=\frac{1+\mathrm{sin}\theta +1-\mathrm{sin}\theta }{1+\mathrm{sin}\theta }=\frac{2}{1+\mathrm{sin}\theta }$
$f\left(x\right)=f\left(x+T\right)$
$f\left(x\right)=\frac{2}{1+\mathrm{sin}\theta }$
$\frac{2}{1+\mathrm{sin}\theta }=\frac{2}{1+\mathrm{sin}\left(\theta +T\right)}\to \mathrm{sin}\left(\theta \right)=\mathrm{sin}\left(\theta +T\right)$
$T+\theta =\theta +2k\pi \to t=2k\pi \to T=2\pi$

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