$\vec{x},\vec{y},\vec{z}$ are the unit vectors along the bisectors of the angles $\alpha,\beta,\gamma$ respectively. If $\vec{u},\vec{v},\vec{w}$ are three non-coplanar unit vectors and $\alpha,\beta,\gamma$ are the angles between $\vec{u}$ and $\vec{v},\vec{v}$ and $\vec{w},$ $\vec{w}$ and $\vec{u}$ respectively and $\vec{x},\vec{y},\vec{z}$ are the unit vectors along the bisectors of the angles $\alpha,\beta,\gamma$ respectively.Prove that $[\vec{x}\times\vec{y}\hspace{0.5cm}\vec{y}\times\vec{z}\hspace{0.5cm}\vec{z}\times\vec{x}]=\frac{1}{16}[\vec{u}\hspace{0.5cm}\vec{v}\hspace{0.5cm}\vec{w}]^2\sec^2\frac{\alpha}{2}\sec^2\frac{\beta}{2}\sec^2\frac{\gamma}{2}$

As $\vec{x},\vec{y},\vec{z},\vec{u},\vec{v},\vec{w},$ are all unit vectors
And $\vec{x}$ is the angle bisector of $\vec{u}$ and $\vec{v}$
$\vec{x}=\vec{u}+\vec{v}$
Similarly,$\vec{y}=\vec{v}+\vec{w}$
Similarly,$\vec{z}=\vec{w}+\vec{u}$
I found $[\vec{x}\times\vec{y}\hspace{0.5cm}\vec{y}\times\vec{z}\hspace{0.5cm}\vec{z}\times\vec{x}]=4[\vec{u}\hspace{0.5cm}\vec{v}\hspace{0.5cm}\vec{w}]^2$
But i cannot get the required answer.Please help me.

The angle bisector between two vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{r}=t(\hat{a}+\hat{b})$ and not $t(\vec{a}+\vec{b})$
So in this case we let $$\vec{x}=\lambda(\hat{u}+\hat{v})$$
$$\vec{y}=\mu(\hat{v}+\hat{w})$$
$$\vec{z}=\Gamma(\hat{w}+\hat{u})$$
Squaring both sides in each case, we obtain:
$$\lambda^2= \frac 14\sec^2(\alpha/2)$$
$$\mu^2=\frac 14 \sec^2(\beta/2) $$
$$\Gamma^2=\frac 14\sec^2(\gamma/2 )$$
Then we have $$[\vec{x}\times\vec{y}\hspace{0.5cm}\vec{y}\times\vec{z}\hspace{0.5cm}\vec{z}\times\vec{x}]=[\vec{x}\space\space\vec{y}\space\space\vec{z}]^2$$
Now put in the values of the vectors $\vec{x},\vec{y}$ and $\vec{z}$ and manipulate the scalar triple product to obtain the desired expression.
