+0

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

+2
269
1
+295

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
+5935
+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
Sort:

#1
+5935
+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

### 10 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details