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}