If $\theta \in\left(0, \frac{\pi}{2}\right)$, then \(\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…

Question

If $\theta \in\left(0, \frac{\pi}{2}\right)$, then $\left|\begin{array}{ccc} (\sin \theta+\operatorname{cosec} \theta)^2 & (\sin \theta-\operatorname{cosec} \theta)^2 & 2020 \ (\cos \theta+\sec \theta)^2 & (\cos \theta-\sec \theta)^2 & 2020 \ (\tan \theta+\cot \theta)^2 & (\tan \theta-\cot \theta)^2 & 2020 \end{array}\right|=$

Aicharya · Accepted Answer

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.

If \(\theta \in\left(0, \frac{\pi}{2}\right)\), then \(\left|\begin{array}{ccc} (\sin…

Solution