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# Suppose that sec(t)=-13/3 and that t is in quadrant 2. Find the exact value of tan(t)?

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Suppose that sec(t)=-13/3 and that t is in quadrant 2. Find the exact value of tan(t).

Deathstroke_rule  Mar 26, 2017
edited by Deathstroke_rule  Mar 26, 2017

#1
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$$\sec(t) = -\frac{13}{3} \\~\\ \cos(t)=-\frac{3}{13}$$

Here is a drawing of that:

By the definition of tangent:

$$\tan (t)=\frac{y}{-3/13}$$

Solve for y with the pythagorean theorem.

$$y^2+(-\frac{3}{13})^2=1 \\~\\ y^2=1-\frac{9}{169} \\~\\ y=\sqrt{\frac{160}{169}} \\~\\ y=\frac{4\sqrt{10}}{13}$$

Now substitute.

$$\tan (t)=\frac{(\frac{4\sqrt{10}}{13})}{-3/13} \\~\\ \tan(t)=(\frac{4\sqrt{10}}{13})*(-\frac{13}{3}) \\~\\ \tan(t)=-\frac{4\sqrt{10}}{3}$$

hectictar  Mar 26, 2017
#1
+7155
+3

$$\sec(t) = -\frac{13}{3} \\~\\ \cos(t)=-\frac{3}{13}$$

Here is a drawing of that:

By the definition of tangent:

$$\tan (t)=\frac{y}{-3/13}$$

Solve for y with the pythagorean theorem.

$$y^2+(-\frac{3}{13})^2=1 \\~\\ y^2=1-\frac{9}{169} \\~\\ y=\sqrt{\frac{160}{169}} \\~\\ y=\frac{4\sqrt{10}}{13}$$

Now substitute.

$$\tan (t)=\frac{(\frac{4\sqrt{10}}{13})}{-3/13} \\~\\ \tan(t)=(\frac{4\sqrt{10}}{13})*(-\frac{13}{3}) \\~\\ \tan(t)=-\frac{4\sqrt{10}}{3}$$

hectictar  Mar 26, 2017