Question

Given $sec\theta =\frac{13}{12}$. Calculate all other trigonometric ratio.

Answer 1

Step 1: Given trigonometric ratio is:

$sec\theta =\frac{13}{12}$,

where, $sec\theta =\frac{Hypotenuse}{Base}$

Step 2: Find the perpendicular, using the Pythagoras theorem.

According to Pythagoras theorem,

${\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+(Base{)}^2$

Put $Hypotenuse=13$ and $Base=12$ in the equation.

${13}^2=Perpendicula{r}^2+{12}^2$

$$\begin{gathered} Perpendicula{r}^2+144=169 \\ Perpendicula{r}^2+144-144=169-144 \\ Perpendicula{r}^2=25 \\ Perpendicular=\sqrt{25} \\ Perpendicular=5 \end{gathered}$$

Step 3: Find other trigonometric ratios.

$$\begin{gathered} \sin \left(\theta \right)=\frac{Perpendicular}{Hypotenuse}=\frac{5}{13} \\ \cos \left(\theta \right)=\frac{Base}{Hypotenuse}=\frac{12}{13} \\ \tan \left(\theta \right)=\frac{Perpendicular}{Base}=\frac{5}{12} \\ \cos ec\left(\theta \right)=\frac{Hypotenuse}{Perpendicular}=\frac{13}{5} \\ cot\left(\theta \right)=\frac{Base}{Perpendicular}=\frac{12}{5} \end{gathered}$$

So, when the value of $sec\left(\theta \right)=\frac{13}{12}$, the values of other trigonometric ratios are $\sin \left(\theta \right)=\frac{5}{13},\cos \left(\theta \right)=\frac{12}{13},\tan \left(\theta \right)=\frac{5}{12},\cos ec\left(\theta \right)=\frac{13}{5},cot\left(\theta \right)=\frac{12}{5}$.