Question

$∆=\left|\begin{array}{ccc}\cos \alpha \cos \mathrm{\beta }& \cos \alpha \sin \mathrm{\beta }& -\sin \alpha \\ -\sin \mathrm{\beta }& \cos \beta & 0\\ \sin \alpha \cos \mathrm{\beta }& \sin \alpha \sin \mathrm{\beta }& \cos \alpha \end{array}\right|$

Answer 1

Given: $∆=\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right|$


$$\begin{gathered} \Rightarrow ∆={\left(-1\right)}^{1+1}\cos \alpha \cos \beta \left(\cos \alpha \cos \beta -0\right)+{\left(-1\right)}^{1+2}\cos \alpha \sin \beta \left(-\sin \beta \cos \alpha -0\right)+{\left(-1\right)}^{1+3}\left(-\sin \alpha \right)\left(-{\sin }^2\beta \sin \alpha -\sin \alpha {\cos }^2\beta \right)\left[\mathrm{Expanding}\mathrm{along}{R}_1\right] \\ =\cos \alpha \cos \beta \left(\cos \alpha \cos \beta -0\right)-\cos \alpha \sin \beta \left(-\sin \beta \cos \alpha -0\right)-\sin \alpha \left(-{\sin }^2\beta \sin \alpha -\sin \alpha {\cos }^2\beta \right) \\ ={\cos }^2\alpha {\cos }^2\beta +{\cos }^2\alpha {\sin }^2\beta +{\sin }^2\alpha {\sin }^2\beta +{\sin }^2\alpha {\cos }^2\beta \\ ={\cos }^2\alpha \left({\cos }^2\beta +{\sin }^2\beta \right)+{\sin }^2\alpha \left({\sin }^2\beta +{\cos }^2\beta \right) \\ \Rightarrow ∆={\cos }^2\alpha +{\sin }^2\alpha \left[\because {\sin }^2\theta +{\cos }^2\theta =1\right] \\ \Rightarrow ∆=1\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right] \end{gathered}$$