Answer :
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
=\cfrac{y}{r}
=\cfrac{1}{csc(\theta)}
\\ \quad \\\\
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
=\cfrac{x}{r}
=\cfrac{1}{sec(\theta)}
\\ \quad \\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
=\cfrac{y}{x}
=\cfrac{sin(\theta)}{cos(\theta)}[/tex]
[tex]\bf cot(\theta)=\cfrac{adjacent}{opposite} =\cfrac{x}{y} =\cfrac{cos(\theta)}{sin(\theta)} \\ \quad \\\\ % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} =\cfrac{r}{y} =\cfrac{1}{sin(\theta)} \\ \quad \\\\ % secant sec(\theta)=\cfrac{hypotenuse}{adjacent} =\cfrac{r}{x} =\cfrac{1}{cos(\theta)}\\\\ -------------------------------\\\\ P(x,y)\implies \delta\qquad tan(\delta)=\cfrac{y}{x}[/tex]
[tex]\bf cot(\theta)=\cfrac{adjacent}{opposite} =\cfrac{x}{y} =\cfrac{cos(\theta)}{sin(\theta)} \\ \quad \\\\ % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} =\cfrac{r}{y} =\cfrac{1}{sin(\theta)} \\ \quad \\\\ % secant sec(\theta)=\cfrac{hypotenuse}{adjacent} =\cfrac{r}{x} =\cfrac{1}{cos(\theta)}\\\\ -------------------------------\\\\ P(x,y)\implies \delta\qquad tan(\delta)=\cfrac{y}{x}[/tex]