+295 Suppose that sec(t)=-13/3 and that t is in quadrant 2. Find the exact value of tan(t).
+9479 \(\sec(t) = -\frac{13}{3} \\~\\ \cos(t)=-\frac{3}{13}\)
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.
\(\tan (t)=\frac{(\frac{4\sqrt{10}}{13})}{-3/13} \\~\\ \tan(t)=(\frac{4\sqrt{10}}{13})*(-\frac{13}{3}) \\~\\ \tan(t)=-\frac{4\sqrt{10}}{3}\)
+9479 \(\sec(t) = -\frac{13}{3} \\~\\ \cos(t)=-\frac{3}{13}\)
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.
\(\tan (t)=\frac{(\frac{4\sqrt{10}}{13})}{-3/13} \\~\\ \tan(t)=(\frac{4\sqrt{10}}{13})*(-\frac{13}{3}) \\~\\ \tan(t)=-\frac{4\sqrt{10}}{3}\)
