allhvasstH

2021-06-12

$\mathrm{tan}\left(x\right)\mathrm{sin}\left(x\right)+\mathrm{sec}\left(x\right){\mathrm{cos}}^{2}\left(x\right)=\mathrm{sec}\left(x\right)$

### Answer & Explanation

Nicole Conner

$\mathrm{tan}\left(x\right)=\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}.$
$\mathrm{tan}\left(x\right)\mathrm{sin}\left(x\right)=\frac{{\mathrm{sin}}^{2}\left(x\right)}{\mathrm{cos}\left(x\right)}.$
$\mathrm{sec}\left(x\right)=\frac{1}{\mathrm{cos}\left(x\right)};$
$\mathrm{sec}\left(x\right){\mathrm{cos}}^{2}\left(x\right)=\mathrm{cos}\left(x\right).$
$\mathrm{tan}\left(x\right)\mathrm{sin}\left(x\right)+\mathrm{sec}\left(x\right){\mathrm{cos}}^{2}\left(x\right)=\frac{{\mathrm{sin}}^{2}\left(x\right)}{\mathrm{cos}\left(x\right)}+\mathrm{cos}\left(x\right)$

$=\frac{{\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)}{\mathrm{cos}\left(x\right)}=\frac{1}{\mathrm{cos}\left(x\right)}=\mathrm{sec}\left(x\right)\left\{proved\right\}$
as we know ${\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)=1$.

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