Question

In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
$sec\theta =\frac{13}{5}$

Answer 1

Given,

$sec\theta =\frac{13}{5}$

$\therefore$ $cos\theta =\frac{1}{sec\theta }=\frac{1}{\frac{13}{5}}=\frac{5}{13}$

We know that,

$\cos \theta =\frac{adjacentSide}{Hypotenuse}$

Consider the figure above,

From Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

${13}^2=5^2+B{C}^2$

$B{C}^2=169-25=144$

$BC=12$

$\sin \theta =\frac{AdjacentSide}{Hypotenuse}=\frac{12}{13}$

$\csc \theta =\frac{1}{\sin \theta }=\frac{1}{\frac{12}{13}}=\frac{13}{12}$

$\tan \theta =\frac{OppositeSide}{AdjacentSide}=\frac{12}{5}$

$\cot \theta =\frac{1}{\tan \theta }=\frac{1}{\frac{12}{5}}=\frac{5}{12}$