Question

Let

${\Delta }_1=\left|\begin{array}{ccc}1& \cos \alpha & \cos \beta \\ \cos \alpha & 1& \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|$

and ${\Delta }_2=\left|\begin{array}{ccc}0& \cos \alpha & \cos \beta \\ \cos \alpha & 0& \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$ .

If ${\Delta }_1={\Delta }_2$, find ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma$

Answer 1

$Given{\Delta }_1=\left|\begin{array}{ccc}1& \cos \alpha & \cos \beta \\ \cos \alpha & 1& \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|$

$expandingthedeterminantgives$

${\Delta }_1={\sin }^2\gamma -{\cos }^2\alpha -{\cos }^2\beta +2\cos \alpha \cos \beta \cos \gamma$

${\Delta }_2=\left|\begin{array}{ccc}0& \cos \alpha & \cos \beta \\ \cos \alpha & 0& \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$

${\Delta }_2=2\cos \alpha \cos \beta \cos \gamma$

$but{\Delta }_1={\Delta }_2$

$\Rightarrow {\sin }^2\gamma -{\cos }^2\alpha -{\cos }^2\beta =0$

$\therefore {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$ [$\because {\cos }^2\alpha =1-{\sin }^2\alpha$]