17Given:
The smaller triangle is,
Find the length of the hypotenuse,
[tex]\begin{gathered} \text{hypotenuse}^2=4^2+1^2 \\ =16+1 \\ =17 \\ \text{hypotenuse}=\sqrt[]{17} \end{gathered}[/tex]The larger traingle is,
[tex]\begin{gathered} \text{Hypotenuse}^2=12^2+3^2 \\ =144+9 \\ =153 \\ \text{Hypotenuse}=3\sqrt[]{17} \end{gathered}[/tex]Now, find all the trigonometric functions for a smaller triangle.
[tex]\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}}=\frac{1}{\sqrt[]{17}} \\ \cos \theta=\frac{Adjacent\text{ side}}{\text{hypotenuse}}=\frac{4}{\sqrt[]{17}} \\ \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}=\frac{1}{4} \\ co\sec \theta=\frac{\text{hypotenuse}}{opposite\text{ side}}=\frac{\sqrt[]{17}}{1}=\sqrt[]{17} \\ \sec \theta=\frac{\text{hypotenuse}}{adjacent\text{ side}}=\frac{\sqrt[]{17}}{4} \\ \cot \theta=\frac{adjecent\text{ side}}{\text{opposite side}}=\frac{4}{1}=4 \end{gathered}[/tex]For the larger triangle,
[tex]\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}}=\frac{3}{3\sqrt[]{17}}=\frac{1}{\sqrt[]{17}} \\ \cos \theta=\frac{Adjacent\text{ side}}{\text{hypotenuse}}=\frac{12}{3\sqrt[]{17}}=\frac{4}{\sqrt[]{17}} \\ \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}=\frac{3}{12}=\frac{1}{4} \\ co\sec \theta=\frac{\text{hypotenuse}}{opposite\text{ side}}=\frac{3\sqrt[]{17}}{3}=\sqrt[]{17} \\ \sec \theta=\frac{\text{hypotenuse}}{adjacent\text{ side}}=\frac{3\sqrt[]{17}}{12}=\frac{\sqrt[]{17}}{4} \\ \cot \theta=\frac{adjecent\text{ side}}{\text{opposite side}}=\frac{12}{3}=4 \end{gathered}[/tex]Therefore, the value of the function is the same because the triangles are similar so, corresponding sides are proportional.
