Question

The question is in picture but for the picture on the bottom can someone tell me if I’m correct or not

Answer 1

`\begin{array}l sin(\theta)=\cfrac{opposite}{hypotenuse} =\cfrac{y}{r} =\cfrac{1}{csc(\theta)}\qquad \\\\\\ cos(\theta)=\cfrac{adjacent}{hypotenuse} =\cfrac{x}{r} =\cfrac{1}{sec(\theta)} \\\\\\ tan(\theta)=\cfrac{opposite}{adjacent} =\cfrac{y}{x} =\cfrac{sin(\theta)}{cos(\theta)} \end{array} ~\hfill~`
`\begin{array}{llll} cot(\theta)=\cfrac{adjacent}{opposite} =\cfrac{x}{y} =\cfrac{cos(\theta)}{sin(\theta)} \\\\\\ csc(\theta)=\cfrac{hypotenuse}{opposite} =\cfrac{r}{y} =\cfrac{1}{sin(\theta)} \\\\\\ sec(\theta)=\cfrac{hypotenuse}{adjacent} =\cfrac{r}{x} =\cfrac{1}{cos(\theta)} \end{array}`