Question

If $\tan \left(x\right)=\frac{5}{12}$ in the third quadrant, then find the other trigonometric ratios.

Answer 1

Step 1: Find the Hypotenuse using right angled triangle approach

We know that in the third quadrant $\mathrm{sine},\mathrm{cosine},\mathrm{secent}$ and $\mathrm{cosecant}$ is negative and $\mathrm{tangent},\mathrm{cotangent}$ is positive.

Given that $\tan \left(x\right)=\frac{5}{12}$ in the third quadrant

We know that $\tan \left(\theta \right)=\frac{\mathrm{Perpendicular}}{\mathrm{Base}}$

$\therefore$ Let $\mathrm{Perpendicular}=5t$ and $\mathrm{Base}=12t$

Therefore the $\mathrm{Hypoteneus}$ is

$$\begin{gathered} \mathrm{Hypoteneus}=\pm \sqrt{{\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perependicular}\right)}^2} \\ \Rightarrow \mathrm{Hypoteneus}=\pm \sqrt{{\left(12t\right)}^2+{\left(5t\right)}^2} \\ \Rightarrow \mathrm{Hypoteneus}=\pm \sqrt{144t^2+25t^2} \\ \Rightarrow \mathrm{Hypoteneus}=\pm \sqrt{169t^2} \\ \Rightarrow \mathrm{Hypoteneus}=\pm 13t \end{gathered}$$

Step 2: Find the value of trigonometric ratios

$$\begin{gathered} \sin \left(x\right)=-\frac{\mathrm{Perpendicular}}{\mathrm{Hypoteneus}}\left[\mathrm{negative}\mathrm{sign}\mathrm{because}x\mathrm{lie}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant}\right] \\ \Rightarrow \sin \left(x\right)=-\frac{5t}{13t} \\ \Rightarrow \sin \left(x\right)=-\frac{5}{13} \end{gathered}$$

$$\begin{gathered} \mathrm{cosec}\left(x\right)=\frac{1}{\sin \left(\mathrm{x}\right)} \\ \Rightarrow \mathrm{cosec}\left(\mathrm{x}\right)=-\frac{13}{5} \end{gathered}$$

$$\begin{gathered} \cos \left(x\right)=-\frac{\mathrm{Base}}{\mathrm{Hypoteneus}}\left[\mathrm{Negative}\mathrm{sign}\mathrm{as}x\mathrm{lie}\mathrm{in}\mathrm{third}\mathrm{quadrant}\right] \\ \Rightarrow \cos \left(x\right)=-\frac{12t}{13t} \\ \Rightarrow \cos \left(x\right)=-\frac{12}{13} \end{gathered}$$

$$\begin{gathered} \sec \left(x\right)=\frac{1}{\cos \left(\mathrm{x}\right)} \\ \Rightarrow \sec \left(\mathrm{x}\right)=-\frac{13}{12} \end{gathered}$$

$$\begin{gathered} cot\left(x\right)=\frac{1}{\tan \left(x\right)} \\ \Rightarrow cot\left(x\right)=\frac{12}{5} \end{gathered}$$

Hence, $\sin \left(x\right)=-\frac{5}{13},\cos \left(x\right)=-\frac{12}{13},\cot \left(x\right)=\frac{12}{5},\sec \left(x\right)=-\frac{13}{12}\mathrm{and}\cos ec\left(x\right)=-\frac{13}{5}$