Question

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

Answer 1

Consider a right-angle triangle  \( Δ \) ABC, right-angled at point B

\(\text{ sec}\ \theta  = \frac{\text{Hypotenuse}}{\text{Side}\ \text{ Adjacent}\ \text{ to}\ ∠\theta }\)

\(=\frac{13}{12}=\frac{AC}{AB}\)
If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in \(Δ\)ABC, we obtain 
\((\text{AC})^ 2 = (\text{AB}) ^2 + (\text{BC})^ 2 \)
\((13k) ^2 = (12k) ^2 + (BC)^ 2 \)
\(169k^ 2 = 144k^ 2 +\text{ BC}^ 2 \)
\(25k ^2 =\text{ BC}^ 2 \)
\(\text{BC} = 5k\)

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

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

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

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