Answer
$\sin{\theta} =\dfrac{n}{\sqrt{m^2+n^2}} $
$\cos{\theta} =\dfrac{m}{\sqrt{m^2+n^2}} $
$\tan{\theta} = \dfrac{n}{m}$
$\csc{\theta} =\dfrac{\sqrt{m^2+n^2}}{n}$
$\sec{\theta} = \dfrac{\sqrt{m^2+n^2}}{m}$
$\cot{\theta} =\dfrac{m}{n}$
Work Step by Step
$\cot{\theta} = \dfrac{x}{y} = \dfrac{m}{n}$
$\therefore x = m \hspace{20pt} y = n $
$\because x$ and $y$ are both positive
$\therefore $The terminal side of $\theta$ in standard position lies in QI
$r = \sqrt{x^2+y^2} = \sqrt{m^2+b^2} $
$\sin{\theta} = \dfrac{y}{r} = \dfrac{n}{\sqrt{m^2+n^2}} $
$\cos{\theta} = \dfrac{x}{r} = \dfrac{m}{\sqrt{m^2+n^2}} $
$\tan{\theta} = \dfrac{y}{x} = \dfrac{n}{m}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = \dfrac{\sqrt{m^2+n^2}}{n}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{\sqrt{m^2+n^2}}{m}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = \dfrac{m}{n}$
