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

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

Guest 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!

Dimitristhym  Oct 19, 2018
edited by Dimitristhym  Oct 19, 2018