Question

If $\mathrm{sec\theta }=\frac{13}{12}$ , find the values of other trigonometric ratios.

Answer 1

$\sec \theta =\frac{13}{12}$ (Given)

$\therefore \cos \theta =\frac{1}{\sec \theta }=\frac{1}{{\displaystyle \frac{13}{12}}}=\frac{12}{13}$
We have,

$$\begin{gathered} 1+{\tan }^2\theta ={\sec }^2\theta \\ \Rightarrow \tan \theta =\sqrt{{\sec }^2\theta -1} \\ \Rightarrow \tan \theta =\sqrt{{\left(\frac{13}{12}\right)}^2-1} \\ \Rightarrow \tan \theta =\sqrt{\frac{169}{144}-1} \end{gathered}$$
$$\begin{gathered} \Rightarrow \tan \theta =\sqrt{\frac{169-144}{144}}=\sqrt{\frac{25}{144}} \\ \Rightarrow \tan \theta =\frac{5}{12} \\ \therefore \cot \theta =\frac{1}{\tan \theta }=\frac{1}{{\displaystyle \frac{5}{12}}}=\frac{12}{5} \end{gathered}$$
Now,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \Rightarrow \sin \theta =\tan \theta \times \cos \theta \\ \Rightarrow \sin \theta =\frac{5}{12}\times \frac{12}{13}=\frac{5}{13} \\ \therefore \mathrm{cosec}\theta =\frac{1}{\sin \theta }=\frac{1}{{\displaystyle \frac{5}{13}}}=\frac{13}{5} \end{gathered}$$