Question

Let $\mathrm{\theta }$ be an angle in quadrant $\mathrm{IV}$ such that $s\mathrm{in\theta }=-\frac{3}{4}$. Find the exact values of $\mathrm{sec\theta }$ and $\mathrm{cot\theta }$.

Answer 1

Step 1: Value of $\mathrm{sec\theta }$:

Given that $\mathrm{\theta }$ be an angle in quadrant $\mathrm{IV}$ and $s\mathrm{in\theta }=-\frac{3}{4}$ .

Apply the identity ${\cos }^2\theta +{\sin }^2\theta =1$.

Substitute $s\mathrm{in\theta }=-\frac{3}{4}$ in the above identity.

$$\begin{gathered} {\cos }^2\mathrm{\theta }+{\left(\frac{-3}{4}\right)}^2=1 \\ \Rightarrow {\cos }^2\mathrm{\theta }+\frac{9}{16}=1 \\ \Rightarrow {\cos }^2\mathrm{\theta }=1-\frac{9}{16} \\ \Rightarrow {\cos }^2\mathrm{\theta }=\frac{16-9}{16} \\ \Rightarrow {\cos }^2\mathrm{\theta }=\frac{7}{16} \end{gathered}$$

Take square root on both sides;

$\mathrm{cos\theta }=\pm \frac{\sqrt{7}}{4}$

In fourth quadrant $c\mathrm{os\theta }$ is positive so, $\mathrm{cos\theta }=\frac{\sqrt{7}}{4}$.

Apply trigonometric identity $\mathrm{sec\theta }=\frac{1}{\mathrm{cos\theta }}$;

Substitute the value of $\mathrm{cos\theta }$.

$\begin{array}{rcl}\mathrm{sec\theta }& =& \frac{1}{{\displaystyle \frac{\sqrt{7}}{4}}}\\ \therefore \mathrm{sec\theta }& =& \frac{4}{\sqrt{7}}\end{array}$

Step 2: Value of $\cot \theta$:

Apply trigonometric identity $\mathrm{cot\theta }=\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}$;

Substitute the value of $\mathrm{cos\theta }\mathrm{and}\mathrm{sin\theta }$.

.$$\begin{gathered} \mathrm{cot\theta }=\frac{{\displaystyle \frac{\sqrt{7}}{4}}}{{\displaystyle \frac{-3}{4}}} \\ \mathrm{cot\theta }=-\frac{\sqrt{7}}{3} \end{gathered}$$;

Hence, the exact values of $\mathrm{sec\theta }\mathrm{is}\frac{4}{\sqrt{7}}$ and $\mathrm{cot\theta }$ is $-\frac{\sqrt{7}}{3}$.