Answer
$\sin{\theta} =\dfrac{a}{\sqrt{a^2+b^2}} $
$\cos{\theta} = \dfrac{b}{\sqrt{a^2+b^2}} $
$\tan{\theta} =\dfrac{a}{b}$
$\csc{\theta} =\dfrac{\sqrt{a^2+b^2}}{a}$
$\sec{\theta} = \dfrac{\sqrt{a^2+b^2}}{b}$
$\cot{\theta} = \dfrac{b}{a}$
Work Step by Step
$\tan{\theta} = \dfrac{y}{x} = \dfrac{a}{b}$
$\therefore y = a \hspace{20pt} x = b $
$\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{b^2+a^2} $
$\sin{\theta} = \dfrac{y}{r} = \dfrac{a}{\sqrt{a^2+b^2}} $
$\cos{\theta} = \dfrac{x}{r} = \dfrac{b}{\sqrt{a^2+b^2}} $
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = \dfrac{\sqrt{a^2+b^2}}{a}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{\sqrt{a^2+b^2}}{b}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = \dfrac{b}{a}$
