Question

Find the values of the trigonometric functions of $\theta$ from the information given.

$\tan \theta =-\frac{3}{4},\cos \theta >0$.

Answer 1

Given:

$\tan \theta =-\frac{3}{4},\cos \theta >0$

To find:

The values of the trigonometric functions of $\theta$.

Explanation:

To find the value of $cot\theta$ we have $cot\theta =\frac{1}{\tan \theta }$.

Substitute $\tan \theta$ as $-\frac{3}{4}$ in the above equation.

$\begin{array}{rcl}cot\theta & =& \frac{1}{-{\displaystyle \frac{3}{4}}}\\ & =& -\frac{4}{3}\end{array}$

Therefore, the value of $cot\theta$ is $-\frac{4}{3}$.

We know that, $se{c}^2\theta =1+{\tan }^2\theta$.

Now substitute the value of $\tan \theta$ as $-\frac{3}{4}$ in the above trigonometric formula.

$\begin{array}{rcl}se{c}^2\theta & =& 1+{\left(-\frac{3}{4}\right)}^2\\ & =& 1+\frac{9}{16}\\ & =& \frac{16+9}{16}\\ & =& \frac{25}{16}\end{array}$

Now square root on both sides of the equation

$\begin{array}{rcl}\sqrt{se{c}^2\theta }& =& \sqrt{\frac{25}{16}}\\ sec\theta & =& \frac{5}{4}\end{array}$

Therefore, the value of $sec\theta$ is $\frac{5}{4}$.

Now we know that, $\cos \theta =\frac{1}{sec\theta }$.

Substitute the value of $sec\theta$ as $\frac{5}{4}$ in the above equation.

$\begin{array}{rcl}\cos \theta & =& \frac{1}{{\displaystyle \frac{5}{4}}}\\ & =& \frac{4}{5}>0\end{array}$

Therefore, the value of $\cos \theta$ is $\frac{4}{5}$.

Now to find the value of $\sin \theta$, we have $\tan \theta =\frac{\sin \theta }{\cos \theta }$.

Substitute the values of $\cos \theta$ as $\frac{4}{5}$ and $\tan \theta$ as $-\frac{3}{4}$ in the above equation.

$\begin{array}{rcl}-\frac{3}{4}& =& \frac{\sin \theta }{{\displaystyle \frac{4}{5}}}\\ -\frac{3}{4}\times \frac{4}{5}& =& \sin \theta \\ \sin \theta & =& -\frac{3}{5}\end{array}$

Therefore, the value of $\sin \theta$ is $-\frac{3}{5}$.

To find the value of $\cos ec\theta$, we know that $\cos ec\theta =\frac{1}{\sin \theta }$.

Substitute the values of $\sin \theta$ as $-\frac{3}{5}$ in the above equation.

$\begin{array}{rcl}\cos ec\theta & =& \frac{1}{-{\displaystyle \frac{3}{5}}}\\ & =& -\frac{5}{3}\end{array}$

Therefore, the values of the trigonometric functions of $\theta$ are

$$\begin{gathered} \sin \theta =-\frac{3}{5},\cos \theta =\frac{4}{5},\tan \theta =-\frac{3}{4}, \\ cot\theta =-\frac{4}{3},sec\theta =\frac{5}{4},\cos ec\theta =-\frac{5}{3} \end{gathered}$$