Answer
$\sin{\theta} =\dfrac{\sqrt{b^2-1}}{b}$
$\cos{\theta} =\dfrac{1}{b}$
$\tan{\theta} =\sqrt{b^2-1}$
$\csc{\theta} =\dfrac{b}{\sqrt{b^2-1}}$
$\sec{\theta} = b$
$\cot{\theta} =\dfrac{1}{\sqrt{b^2-1} }$
Work Step by Step
$\sec{\theta} = b$
$\cos{\theta} = \dfrac{1}{\sec{\theta}} = \dfrac{1}{b}$
$\because \theta \in QI \hspace{20pt} \therefore \sin{\theta}$ is positive.
$\sin{\theta} = \sqrt{1-\cos^2{\theta}} = - \sqrt{1-\left(\dfrac{1}{b}\right)^2} = \sqrt{\dfrac{b^2-1}{b^2}} = \dfrac{\sqrt{b^2-1}}{b}$
$\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} = \sqrt{b^2-1}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = \dfrac{b}{\sqrt{b^2-1}}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}= \dfrac{1}{\sqrt{b^2-1}} $
