Question

Find the values of the trigonometric functions of $t$from the given information.

$\csc t=5,\cos t<0$

Short answer

The values of the trigonometric functions are $\mathbf{cos}\mathit{t}\mathbf{=}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{,}\mathbf{tan}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{,}\mathbf{sin}\mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{,}$$\mathbf{sec}\mathit{t}\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}}$, and $\mathbf{cot}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}$.

Step-by-step solution

Step 1. State the Trigonometric identities.

The reciprocal identities are $\csc t=\frac{1}{\sin t},\sec t=\frac{1}{\cos t},\cot t=\frac{1}{\tan t},\tan t=\frac{\sin t}{\cos t}$and $\cot =\frac{\cos t}{\sin t}$.

The Pythagorean identities are:

1. ${\sin }^{2}t+{\cos }^{2}t=1$

2. ${\tan }^{2}t+1={\sec }^{2}t$

3. ${\cot }^{2}t+1={\csc }^{2}t$

Step 2. Find cost and sint. 

Substitute 5 for $\csc t$into reciprocal identity $\csc t=\frac{1}{\sin t}$and simplify for $\sin t$.

$$\begin{gathered} \csc t=\frac{1}{\sin t} \\ 5=\frac{1}{\sin t} \\ \sin t=\frac{1}{5} \end{gathered}$$

Now use Pythagorean Identity ${\sin }^{2}t+{\cos }^{2}t=1$, substitute $\frac{1}{5}$for $\sin t$and simplify for $\cos t$.

$$\begin{gathered} {\sin }^{2}t+{\cos }^{2}t=1 \\ {\left(\frac{1}{5}\right)}^2+{\cos }^{2}t=1 \end{gathered}$$

$$\begin{gathered} \frac{1}{25}+{\cos }^{2}t-\frac{1}{25}=1-\frac{1}{25} \\ {\cos }^{2}t=\frac{24}{25} \\ \cos t=\pm \frac{2\sqrt{6}}{5} \end{gathered}$$

Since, $\cos t<0$, therefore, $\cos t=-\frac{2\sqrt{6}}{5}$.

Step 3. Find other trigonometric functions.

Now substitute $\frac{1}{5}$for $\sin t$and $-\frac{2\sqrt{6}}{5}$for into reciprocal identity $\tan t=\frac{\sin t}{\cos t}$and simplify for $\tan t$.

$$\begin{gathered} \tan t=\frac{\sin t}{\cos t} \\ =\frac{\left(\frac{1}{5}\right)}{\left(-\frac{2\sqrt{6}}{5}\right)} \\ =-\frac{\sqrt{6}}{12} \end{gathered}$$

$$\begin{gathered} \sec t=\frac{1}{\cos t} \\ =\frac{1}{\left(-\frac{2\sqrt{6}}{5}\right)} \\ =-\frac{5\sqrt{6}}{12}\left(\text{Rationalise}\right) \end{gathered}$$

$$\begin{gathered} \cot t=\frac{1}{\tan t} \\ =\frac{1}{\left(-\frac{\sqrt{6}}{12}\right)} \\ =-2\sqrt{6} \end{gathered}$$