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If sec(x) = -7/3, and in the second quadrant, find tan(x) and sin(x)

 Oct 19, 2018
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sec(x) = -7/3 <=> \(\frac{1}{cos(x)}=\frac{-7}{3} \) <=> \(cos(x) = \frac{-3}{7}\) 

\(cos^2(x)=\frac{9}{49}\),We know \(sin^2(x) + cos^2(x) =1\)

So \(1-cos^2(x)=\frac{49-9}{49}\)<=> \(sin^2(x)=\frac{40}{49}\) => \(sin(x)=\frac{\sqrt{40}}{7}\) because is in the second quadrant its 

positive

and \(tan(x)=\frac{sin(x)}{cos(x)}=\frac{\frac{\sqrt{40}}{7}}{\frac{-3}{7}}\)<=> \(tan(x)=-\frac{\sqrt{40}}{3}\)

 

Hope it helps! 

 Oct 19, 2018
edited by Dimitristhym  Oct 19, 2018

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