Q. 58 - Precalculus Enhanced with Graphing Utilities

Question

Find the exact value of each of the remaining trigonometric functions of $\mathrm{\theta }.$

$\mathrm{sec\theta }=-2,\mathrm{tan\theta }>0$

Step-by-Step Solution

Answer

The exact values of trigonometric functions are,

$$\begin{gathered} \mathrm{sin\theta }=-\frac{\sqrt{3}}{2} \\ \mathrm{tan\theta }=\sqrt{3} \\ \mathrm{cot\theta }=\frac{\sqrt{3}}{3} \\ \mathrm{csc\theta }=-\frac{2\sqrt{3}}{3} \end{gathered}$$

Step 1. Given Information

Given that the trigonometric function is $\mathrm{sec\theta }=-2,\mathrm{tan\theta }>0.$ To find the exact values of trigonometric function using substitution.

Step 2. Solving a equation

It implies that $\mathrm{sin\theta }<0$ in III quadrant. On using the equation,

$$\begin{gathered} {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1 \\ \mathrm{sin\theta }=\pm \sqrt{1-{\cos }^2\mathrm{\theta }} \end{gathered}$$

On substituting the value of $\mathrm{cos\theta }=-\frac{1}{2}$ in the equation,

$$\begin{gathered} \mathrm{sin\theta }=-\sqrt{1-{\left(-\frac{1}{2}\right)}^2} \\ =-\sqrt{1-\frac{1}{4}} \\ \mathrm{sin\theta }=-\frac{\sqrt{3}}{2} \end{gathered}$$

Step 3. Substitution

On substituting the value in equation,

$$\begin{gathered} \mathrm{tan\theta }=\frac{\mathrm{y}}{\mathrm{x}} \\ =\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }} \\ =\frac{-{\displaystyle \frac{\sqrt{3}}{2}}}{-{\displaystyle \frac{1}{2}}} \\ \mathrm{tan\theta }=\sqrt{3} \\ \mathrm{cot\theta }=\frac{1}{\mathrm{tan\theta }} \\ \mathrm{cot\theta }=\frac{\sqrt{3}}{3} \\ \mathrm{csc\theta }=\frac{1}{\mathrm{sin\theta }} \\ =\frac{1}{-{\displaystyle \frac{\sqrt{3}}{2}}} \\ \mathrm{csc\theta }=-\frac{2\sqrt{3}}{3} \end{gathered}$$