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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

Best Answer 

 #1
avatar+5541 
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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
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1+0 Answers

 #1
avatar+5541 
+3
Best Answer

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

 

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

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