Question

Given sec \( θ =\frac{13}{12}\), calculate all other trigonometric ratios.

Answer 1

Consider a right-angled triangle \( Δ \) ABC, with the right angle at B. Given that sec \( \theta = \frac{\text{Hypotenuse}}{\text{Side Adjacent to } ∠\theta } = \frac{13}{12} = \frac{AC}{AB} \).

Let AC = 13k and AB = 12k, where k is a positive integer.

Applying the Pythagoras theorem to \(Δ\)ABC:

\((\text{AC})^ 2 = (\text{AB}) ^2 + (\text{BC})^ 2 \)

\((13k) ^2 = (12k) ^2 + (\text{BC})^ 2 \)

\(169k^ 2 = 144k^ 2 + (\text{BC})^ 2 \)

\(25k ^2 = (\text{BC})^ 2 \)

\(\text{BC} = 5k\)

The other trigonometric ratios are:

\(\text{ sin}\ \theta = \frac{\text{Side Opposite to } ∠\theta }{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}\)

\(\text{ cos}\ \theta = \frac{\text{Side Adjacent to } ∠\theta }{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}\)

\(\text{tan}\ (\theta) = \frac{\text{Side Opposite to } ∠\theta}{\text{Side Adjacent to } ∠\theta} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}\)

\(\text{cosec} \ (\theta) = \frac{\text{Hypotenuse}}{\text{Side Opposite to } ∠\theta} = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}\)