Respuesta :

sqdancefan
[tex]\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^{2}}}=\dfrac{\sin(\theta)}{\cos(\theta)}\\\\=\tan(\theta)[/tex]
gmany
[tex]\text{Use:}\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\tan x=\dfrac{\sin x}{\cos x}[/tex]

[tex]\dfrac{\sin\Theta}{\sqrt{1-\sin^2\Theta}}=\dfrac{\sin\Theta}{\sqrt{\cos^2\Theta}}=\dfrac{\sin\Theta}{\cos\Theta}=\tan\Theta[/tex]

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