In any triangle $ABC$,$(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2} =$

In $\triangle ABC$, if $\angle A = 90^{\circ}$, then $\sin(B - C) =$

If in a triangle $ABC$,$\angle C = 60^{\circ}$,then $\frac{1}{a + c} + \frac{1}{b + c} = $

In $\triangle ABC$,if $4r_1 = 5r_2 = 6r_3$,then $\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} =$

In $\triangle ABC$, if $\angle C = \frac{\pi}{3}$, then the value of $\cos^2 A + \cos^2 B - \cos A \cos B$ is...

Points $D$ and $E$ are taken on the side $BC$ of a triangle $ABC$ such that $BD = DE = EC$. If $\angle BAD = x$,$\angle DAE = y$,and $\angle EAC = z$,then the value of $\frac{\sin(x + y)\sin(y + z)}{\sin x \sin z}$ is: