Question

find the value of the remaining five trigonometric functions of

$$\begin{gathered} \tan \vartheta =\frac{-12}{5} \\ \pi /2<\vartheta <\pi \end{gathered}$$

Short answer

the value of the remaining five trigonometric functions of $\vartheta$are:

$$\begin{gathered} \cos \vartheta =-\frac{5}{13} \\ \sin \vartheta =\frac{12}{13} \\ \cos ec\vartheta =\frac{13}{12} \\ sec\vartheta =-\frac{13}{5} \\ cot\vartheta =-\frac{5}{12} \end{gathered}$$

Step-by-step solution

Step 1. Given information

$$\begin{gathered} \tan \vartheta =\frac{-12}{5} \\ \pi /2<\vartheta <\pi \end{gathered}$$

Step 2. Solving for  secϑ & cosϑ

We know that ;

${\tan }^2\vartheta +1=se{c}^2\vartheta$

Using this equation:

$$\begin{gathered} sec\vartheta =\sqrt{1+{\tan }^2\vartheta } \\ sec\vartheta =\sqrt{1+{(-\frac{12}{5})}^2} \\ sec\vartheta =\sqrt{1+\frac{144}{25}} \\ sec\vartheta =\sqrt{\frac{169}{25}} \\ sec\vartheta =\frac{13}{5} \end{gathered}$$

Since theta is in second quadrant so;

$sec\vartheta =-\frac{13}{5}$

Step 3. Solving forcosϑ , sinϑ , cotϑ ,cosecϑ

We know that :

$$\begin{gathered} \cos ec\vartheta =\frac{1}{\sin \vartheta } \\ sec\vartheta =\frac{1}{\cos \vartheta } \\ cot\vartheta =\frac{1}{\tan \vartheta } \end{gathered}$$

Using this :

$$\begin{gathered} \cos \vartheta \frac{-5}{13} \\ \sin \vartheta =\cos \vartheta \times \tan \vartheta \\ \sin \vartheta =\frac{12}{13} \\ cot\vartheta =\frac{-5}{12} \\ \cos ec\vartheta =\frac{13}{12} \end{gathered}$$