Science Of Sines

From Direction Cosine Theorem, we get $l^2 + m^2 + n^2 = 1. \ _\square$ , where $l=\cos a ,m=\cos b$ and $n=\cos c$ .

Using $1-{ l }^{ 2 }=\sin ^{ 2 }{ a } \\ 1-{ m }^{ 2 }=\sin ^{ 2 }{ b } \\ 1-{ n }^{ 2 }=\sin ^{ 2 }{ c }$

We get $\sin ^{ 2 }{ a } +\sin ^{ 2 }{b}+ { \sin ^{2 }{ c } } =2$ .

We know, (cos(a))^2+(cos(b))^2+(cos(g))^2=1 => 3-( squares of sine of a,b,c)=1 =>(sin(a))^2+(sin(b))^2+(sin(g))^2=2

It doesn't really proof,but if you assume that this is always true no matter the position of the line,you can just make the calculation based on the line being one of the axes.Thus,the case that the angles are 0,90,90