With usual notations,in a triangle $ABC$,$a \cos(B - C) + b \cos(C - A) + c \cos(A - B)$ is equal to

In a triangle $ABC$,with usual notations $\angle A=60^{\circ}$,then $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

With usual notations in $\triangle ABC$, if $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = \frac{3a}{2}$, then:

Let $p_1, p_2, p_3$ be the altitudes of a triangle $ABC$ drawn through the vertices $A, B, C$ respectively. If $r_1=4, r_2=6, r_3=12$ are the ex-radii of triangle $ABC$, then $\frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=$

Let a triangle $ABC$ be inscribed in a circle of radius $2$ units. If the $3$ bisectors of the angles $A, B$ and $C$ are extended to cut the circle at $A_1, B_1$ and $C_1$ respectively,then the value of $\left[\frac{AA_1 \cos \frac{A}{2} + BB_1 \cos \frac{B}{2} + CC_1 \cos \frac{C}{2}}{\sin A + \sin B + \sin C}\right]^2$ is:

If in $\triangle ABC$,with usual notations,$a^2, b^2, c^2$ are in $A$.$P$.,then $\frac{\sin 3B}{\sin B} =$