Question

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

Answer 1

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

\[\therefore \cos\theta = \frac{1}{\sec\theta} = \frac{1}{\frac{13}{12}} = \frac{12}{13}\]
We have,

\[1 + \tan^2 \theta = \sec^2 \theta\]

\[ \Rightarrow \tan\theta = \sqrt{\sec^2 \theta - 1}\]

\[ \Rightarrow \tan\theta = \sqrt{\left( \frac{13}{12} \right)^2 - 1}\]

\[ \Rightarrow \tan\theta = \sqrt{\frac{169}{144} - 1}\]

\[\Rightarrow \tan\theta = \sqrt{\frac{169 - 144}{144}} = \sqrt{\frac{25}{144}}\]
\[ \Rightarrow \tan\theta = \frac{5}{12}\]
\[ \therefore \cot\theta = \frac{1}{\tan\theta} = \frac{1}{\frac{5}{12}} = \frac{12}{5}\]

Now,

\[\tan\theta = \frac{\sin\theta}{\cos\theta}\]

\[ \Rightarrow \sin\theta = \tan\theta \times \cos\theta\]

\[ \Rightarrow \sin\theta = \frac{5}{12} \times \frac{12}{13} = \frac{5}{13}\]

\[ \therefore cosec\theta = \frac{1}{\sin\theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5}\]