Given determinate, where \(\theta \in\left(0, \frac{\pi}{2}\right)\) is
\(\Delta=\left|\begin{array}{ccc}
(\sin \theta+\operatorname{cosec} \theta)^2 & (\sin \theta-\operatorname{cosec} \theta)^2 & 2020 \\
(\cos \theta+\sec \theta)^2 & (\cos \theta-\sec \theta)^2 & 2020 \\
(\tan \theta+\cot \theta)^2 & (\tan \theta-\cot \theta)^2 & 2020
\end{array}\right|\)
On applying \(C_1 \rightarrow C_1-C_2\), we get
\(\Delta=\left|\begin{array}{ccc}
4 & (\sin \theta-\operatorname{cosec} \theta)^2 & 2020 \\
4 & (\cos \theta-\sec \theta)^2 & 2020 \\
4 & (\tan \theta-\cot \theta)^2 & 2020
\end{array}\right|\)
\(\because\) Elements of columns \(C_1\) and \(C_2\) are perportional, so \(\Delta=0\)
Hence, option (c) is correct.