Answer
$\begin{array}{|c|c|}\hline
\sin\theta = \dfrac{24}{25} \text{(given)} & \color{blue}{\csc\theta = \dfrac{25}{24}} \\ \hline
\color{blue}{\cos\theta = -\dfrac{7}{25}} & \color{blue}{\sec\theta = -\dfrac{25}{7}} \\ \hline
\color{blue}{\tan\theta = -\dfrac{24}{7}} & \color{blue}{\cot\theta = -\dfrac{7}{24}} \\ \hline
\end{array}$
Work Step by Step
Given: $\sin\theta = \frac{24}{25},\ \theta \in {\rm QII}$
Since $\theta \in {\rm QII}$, then $\begin{array}{|c|c|}\hline
\sin\theta \gt 0 & \csc\theta \gt 0 \\ \hline
\cos\theta\lt 0 & \sec\theta \lt 0 \\ \hline
\tan\theta \lt 0 & \cot\theta \lt 0 \\ \hline
\end{array}$.
(i) $\color{blue}{\csc\theta} = \dfrac{1}{\sin\theta} = \dfrac{1}{24/25} = \color{blue}{\dfrac{25}{24}}$.
(ii) From $\cos^2\theta + \sin^2\theta = 1$, we have
$\quad \begin{eqnarray}
\cos^2\theta &=& 1-\sin^2\theta \\
&=& 1 - \left(\frac{24}{25}\right)^2 \\
&=& 1 - \frac{576}{625} \\
&=& \frac{625-576}{625} \\
&=& \frac{49}{625} \\
\cos\theta &=& \pm\sqrt{\frac{49}{625}} \\
&=& \pm \frac{7}{25} \\
\color{blue}{\cos\theta} &=& \color{blue}{-\frac{7}{25}},\quad \text{since}\ \cos\theta \lt 0\ \text{in QII}.
\end{eqnarray}$
(iii) $\color{blue}{\sec\theta} = \dfrac{1}{\cos\theta} = \dfrac{1}{-7/25} = \color{blue}{-\dfrac{25}{7}}$.
(iv) $\color{blue}{\tan\theta} = \dfrac{\sin\theta}{\cos\theta} = \dfrac{24/25}{-7/25} = \color{blue}{-\dfrac{24}{7}}$.
(v) $\color{blue}{\cot\theta} = \dfrac{1}{\tan\theta} = \dfrac{1}{-24/7} = \color{blue}{-\dfrac{7}{24}}$.
Summarizing (i)-(v):
$\begin{array}{|c|c|}\hline
\sin\theta = \dfrac{24}{25} \text{(given)} & \color{blue}{\csc\theta = \dfrac{25}{24}} \\ \hline
\color{blue}{\cos\theta = -\dfrac{7}{25}} & \color{blue}{\sec\theta = -\dfrac{25}{7}} \\ \hline
\color{blue}{\tan\theta = -\dfrac{24}{7}} & \color{blue}{\cot\theta = -\dfrac{7}{24}} \\ \hline
\end{array}$
