Answer
$\begin{array}{|c|c|}\hline
\sin\theta = \dfrac{1}{a} \text{(given)} & \color{blue}{\csc\theta = a} \\ \hline
\color{blue}{\cos\theta = \dfrac{\sqrt{a^2-1}}{a}} & - \\ \hline
- & \color{blue}{\cot\theta =\sqrt{a^2-1}} \\ \hline
\end{array}$
Work Step by Step
Given: $\sin\theta = \frac{1}{a},\ \theta \in {\rm QI}$
Since $\theta \in {\rm QI}$, then $\begin{array}{|c|c|}\hline
\sin\theta \gt 0 & \csc\theta \gt 0 \\ \hline
\cos\theta\gt 0 & \sec\theta \gt 0 \\ \hline
\tan\theta \gt 0 & \cot\theta \gt 0 \\ \hline
\end{array}$.
Since $1 \gt \sin\theta \gt 0$ for $\theta\in\rm QI$, then $1 \gt 1/a \gt 0 \implies a>1$.
(i) $\color{blue}{\csc\theta} = \dfrac{1}{\sin\theta} = \dfrac{1}{1/a} = \color{blue}{a}$.
(ii) Since $\cos^2\theta + \sin^2\theta = 1$, then
$\quad \begin{eqnarray}
\cos^2\theta &=& 1-\sin^2\theta \\
&=& 1 - \left(\frac{1}{a}\right)^2 \\
&=& 1 - \frac{1}{a^2} \\
\cos^2\theta &=& \frac{a^2-1}{a^2} \\
\cos\theta &=& \sqrt{\frac{a^2-1}{a^2}}\quad \text{since}\ \theta\in{\rm QI} \\
\cos\theta &=& \frac{\sqrt{a^2-1}}{\sqrt{a^2}}\\
\color{blue}{\cos\theta} &=& \color{blue}{\frac{\sqrt{a^2-1}}{a}},\quad \text{since}\ \sqrt{a^2} = |a|= a\quad \text{as } a>0.
\end{eqnarray}$
(iii) $\color{blue}{\cot\theta} = \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sqrt{a^2-1}/a}{1/a} = \color{blue}{\sqrt{a^2-1}}$.
Summarizing (i)-(iii):
$\begin{array}{|c|c|}\hline
\sin\theta = \dfrac{1}{a} \text{(given)} & \color{blue}{\csc\theta = a} \\ \hline
\color{blue}{\cos\theta = \dfrac{\sqrt{a^2-1}}{a}} & - \\ \hline
- & \color{blue}{\cot\theta =\sqrt{a^2-1}} \\ \hline
\end{array}$
