91Ó°ÊÓ

The given figure is shown below.

The six trigonometric functions are $\sin \mathrm{θ},  \csc \mathrm{θ},  \cos \mathrm{θ},  \sec \mathrm{θ},  \tan \mathrm{θ},  \cot \mathrm{θ}$.

Let $(x,y)$ be a point in X-Y plane and (r)be the length of a line segment drawn from the origin to the point.

In, $\mathrm{Δ}OAB,OA=r, OB=x, AB=y$

Therefore, six trigonometric functions are;

$\begin{array}{c}\sin \mathrm{θ}=\frac{AB}{OA}\text{ }=\frac{y}{r}\text{ },\text{ }\csc \mathrm{θ}=\frac{1}{\sin \mathrm{θ}}=\frac{OA}{AB}=\frac{r}{y}\\ \cos \mathrm{θ}=\frac{OB}{OA}=\frac{x}{r},\sec \mathrm{θ}=\frac{1}{\cos \mathrm{θ}}=\frac{OA}{OB}=\frac{r}{x}\\ \tan \mathrm{θ}=\frac{AB}{OB}=\frac{y}{x},\cot \mathrm{θ}=\frac{1}{\tan \mathrm{θ}}=\frac{OB}{AB}=\frac{x}{y}\end{array}$

From the given figure, the hypotenuse is evaluated as:

$\begin{array}{c}\text{Hypotenuse}=\sqrt{8^2+{15}^2}\\ =\sqrt{64+225}\\ =17\end{array}$

Now, the six trigonometric functions are evaluated as:

$\begin{array}{c}\sin \mathrm{θ}=\frac{8}{17}\csc \mathrm{θ}=\frac{17}{8}\\ \cos \mathrm{θ}=\frac{15}{17}\sec \mathrm{θ}=\frac{17}{15}\\ \tan \mathrm{θ}=\frac{8}{15}\cot \mathrm{θ}=\frac{15}{8}\end{array}$

The six trigonometric functions are: $\sin \mathrm{θ}=\frac{8}{17},\csc \mathrm{θ}=\frac{17}{8},\cos \mathrm{θ}=\frac{15}{17},\sec \mathrm{θ}=\frac{17}{15},\tan \mathrm{θ}=\frac{8}{15},\cot \mathrm{θ}=\frac{15}{8}$