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:

$\begin{array}{c}\text{Base}=\sqrt{{12}^2−{10}^2}\\ =2\sqrt{11}\end{array}$

Now, the six trigonometric functions are evaluated as:

localid="1648533344145" $\begin{array}{c}\sin \mathrm{θ}=\frac{10}{12}=\frac{5}{6},\csc \mathrm{θ}=\frac{6}{5}\\ \cos \mathrm{θ}=\frac{2\sqrt{11}}{12}=\frac{\sqrt{11}}{6},\sec \mathrm{θ}=\frac{6}{\sqrt{11}}\\ \tan \mathrm{θ}=\frac{5}{\sqrt{11}},\cot \mathrm{θ}=\frac{\sqrt{11}}{5}\end{array}$

The six trigonometric functions are: $\sin \mathrm{θ}=\frac{5}{6},\csc \mathrm{θ}=\frac{6}{5},\cos \mathrm{θ}=\frac{\sqrt{11}}{6},\sec \mathrm{θ}=\frac{6}{\sqrt{11}},\tan \mathrm{θ}=\frac{5}{\sqrt{11}},\cot \mathrm{θ}=\frac{\sqrt{11}}{5}$