If a line makes angles $\alpha ,\beta ,\gamma ,\delta$ with the 4 diagonals of a cube then ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +{\sin }^2\delta =$

A line makes angles $\alpha ,\beta ,\gamma ,\delta$ with 4 diagonals of a cube

We know that ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{4}{3}$

$1-{\sin }^2\alpha +1-{\sin }^2\beta +1-{\sin }^2\gamma +1-{\sin }^2\delta =\frac{4}{3}$

$4-\frac{4}{3}={\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +{\sin }^2\delta$

$\therefore {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +{\sin }^2\delta =\frac{8}{3}$