If $A_{\alpha }=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$, then determinant of $A_{\pi /5}{A}_{\pi /4}{A}_{3\pi /10}$

Given$A_{\alpha }=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$

$A_{\frac{\mathrm{\pi }}{5}}=\left[\begin{array}{cc}\cos \frac{\mathrm{\pi }}{5}& \sin \frac{\mathrm{\pi }}{5}\\ -\sin \frac{\mathrm{\pi }}{5}& \cos \frac{\mathrm{\pi }}{5}\end{array}\right], A_{\frac{\mathrm{\pi }}{4}}=\left[\begin{array}{cc}\cos \frac{\mathrm{\pi }}{4}& \sin \frac{\mathrm{\pi }}{4}\\ -\sin \frac{\mathrm{\pi }}{4}& \cos \frac{\mathrm{\pi }}{4}\end{array}\right], A_{\frac{3\mathrm{\pi }}{10}}=\left[\begin{array}{cc}\cos \frac{3\mathrm{\pi }}{10}& \sin \frac{3\mathrm{\pi }}{10}\\ -\sin \frac{3\mathrm{\pi }}{10}& \cos \frac{3\mathrm{\pi }}{10}\end{array}\right]$

Determinant of $A_{\pi /5}{A}_{\pi /4}{A}_{3\pi /10}$

$\left|A_{\frac{\mathrm{\pi }}{5}}\right|=\left|\begin{array}{cc}\cos \frac{\mathrm{\pi }}{5}& \sin \frac{\mathrm{\pi }}{5}\\ -\sin \frac{\mathrm{\pi }}{5}& \cos \frac{\mathrm{\pi }}{5}\end{array}\right|$

$={\cos }^2\frac{\mathrm{\pi }}{5}+{\sin }^2\frac{\mathrm{\pi }}{5}$

$=1$

$\left|{\mathrm{A}}_{\frac{\mathrm{\pi }}{4}}\right|=\left|\begin{array}{cc}\cos \frac{\mathrm{\pi }}{4}& \sin \frac{\mathrm{\pi }}{4}\\ -\sin \frac{\mathrm{\pi }}{4}& \cos \frac{\mathrm{\pi }}{4}\end{array}\right|$

$={\cos }^2\frac{\mathrm{\pi }}{4}+{\sin }^2\frac{\mathrm{\pi }}{4}$

$=1$

$\left|A_{\frac{3\mathrm{\pi }}{10}}\right|=\left|\begin{array}{cc}\cos \frac{3\mathrm{\pi }}{10}& \sin \frac{3\mathrm{\pi }}{10}\\ -\sin \frac{3\mathrm{\pi }}{10}& \cos \frac{3\mathrm{\pi }}{10}\end{array}\right|\begin{array}{}\\ \end{array}$

$={\cos }^2\frac{3\mathrm{\pi }}{10}+{\sin }^2\frac{3\mathrm{\pi }}{10}$

$=1$

Value of $A_{\pi /5}{A}_{\pi /4}{A}_{3\pi /10}=1\times 1\times 1=1$