Question

demostre a identidade segunte:

sen alpha tg alpha + cos alpha = sec alpha

Answer 1

Lembre-se:

$sen^{2} \alpha +cos^{2} \alpha =1$
$tg \alpha =sen \alpha /cos \alpha$
$sec \alpha =1/cos \alpha$
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$sen \alpha *tg\alpha+ cos \alpha = sec \alpha$
$sen \alpha * (sen \alpha /cos \alpha ) + cos \alpha = sec \alpha$
$(sen^{2} \alpha /cos \alpha ) + cos \alpha = sec \alpha$
$(sen^{2} \alpha /cos \alpha ) + (cos \alpha *cos \alpha /cos \alpha ) = sec \alpha$
$(sen^{2} \alpha /cos \alpha ) + (cos^{2} \alpha /cos \alpha ) = sec \alpha$
$(sen^{2} \alpha +cos^{2} \alpha )/cos \alpha = sec \alpha$
$1/cos \alpha = sec \alpha$
$sec \alpha =sec \alpha$