` vec u , vec v` and `vec w` are three non-coplanar unit vecrtors and `alpha,beta` and `gamma` are the angles between` vec u` and `vec v , vec v`and `vec w ,a n d vec w` and `vec u ,` respectively, and ` vec x , vec y` and `vec z` are unit vectors along the bisectors of the angles `alpha,betaa n dgamma` , respectively. Prove that `[ vecx xx vec y vec yxx vec z vec zxx vec x]=1/(16)[ vec u vec v vec w]^2sec^2(alpha/2)sec^2(beta/2)sec^2(gamma/2)`.

Explanation

WE know that `[vecx xx vecy vecy xx veczveczxx vecx] = [vecx vecy vecz] ^(2)` Also a vector along the bisector of given two unit vectors
`vecu,vecv,"is " vecu +vecv`
A unit vector along the bisector is `(vecu +vecv)/(|vecu +vecv|)`
`|vecu +vecv|^(2)=1+1+2vecu.vecv=2+2cosalpha=4cos^(2)""alpha/2`
`Rightarrow vecx = (vecu +vecv)/(2cos ""alpha/2)`
similarly , ` vecy= (vecv + vecw)/(2cosbeta//2) and vecz= (vecu+vecw)/(2cosgamma//2)`
`Rightarrow [ vecxvecy vecz] = 1/8[vecu+vecv vecv + vecw vecu + vecw] xx sec "" alpha/2 sec"" beta/2 sec"" gamma/2`
`= 1/82[vecu vecv vecw] sec""alpha/2sec"" beta/2sec"" gamma/2`
`1/4[vecu vecv vecw] sec""alpha/2 sec""beta/2sec""gamma/2`
`Rightarrow [vecx xx vecyvecy xxvecz vecz xx vecx] = [vecx " "vecy " " vecz]^(2)`
`= 1/16[vecu vecv vecw]^(2) sec^(2)""alpha/2 sec^(2)""beta/2 sec^(2)""gamma/2`