# If Sec θ = 13 12 , Find the Values of Other Trigonometric Ratios. - Geometry

#### Question

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

#### Solution

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

$\tan\theta = \frac{\sin\theta}{\cos\theta}$

$\Rightarrow \sin\theta = \tan\theta \times \cos\theta$

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

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

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