Question

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

$\mathrm{cot\theta }=\frac{4}{3},\mathrm{cos\theta }<0.$

Short answer

The exact values for trigonometric functions are,

$$\begin{gathered} \mathrm{sec\theta }=-\frac{5}{4} \\ \mathrm{cos\theta }=-\frac{4}{5} \\ \mathrm{sin\theta }=-\frac{3}{5} \\ \mathrm{csc\theta }=-\frac{5}{3} \end{gathered}$$

Step-by-step solution

Step 1. Given Information

Given that the trigonometric function is $\mathrm{cot\theta }=\frac{4}{3},\mathrm{cos\theta }<0.$To find the exact values of trigonometric functions using substitution.

Step 2. Solve a equation

It implies $\mathrm{sin\theta }<0$where lies in III quadrant. On using the equation of $\mathrm{sec\theta }=\pm \sqrt{{\tan }^2\mathrm{\theta }+1}$.

$$\begin{gathered} \mathrm{sec\theta }=-\sqrt{1+{\left(\frac{3}{4}\right)}^2} \\ =-\sqrt{1+\frac{9}{16}} \\ \mathrm{sec\theta }=-\frac{5}{4} \\ \mathrm{cos\theta }=\frac{1}{\mathrm{sec\theta }} \\ \mathrm{cos\theta }=-\frac{4}{5} \end{gathered}$$

Step 3. Substitution

On using the equation for substitution,

$$\begin{gathered} \mathrm{sin\theta }=\pm \sqrt{1-{\cos }^2\mathrm{\theta }} \\ =-\sqrt{1-{(-\frac{4}{5})}^2} \\ =-\sqrt{1-\frac{16}{25}} \\ \mathrm{sin\theta }=-\frac{3}{5} \\ \mathrm{csc\theta }=\frac{1}{\mathrm{sin\theta }} \\ =\frac{1}{-{\displaystyle \frac{3}{5}}} \\ \mathrm{csc\theta }=-\frac{5}{3} \end{gathered}$$