In $\triangle ABC, (r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} =$

In a triangle,if the ex-radii $r_1, r_2, r_3$ are in the ratio $1: 2: 3$,then its sides are in the ratio

In a $\triangle ABC$,if $a+3b=3c$,then $\sin \frac{A}{2} =$

In $\triangle ABC$,$\angle B=60^{\circ}$ and $\angle A=75^{\circ}$. If a point $D$ divides $BC$ in the ratio $2:3$,then $\sin \angle BAD : \sin \angle CAD=$

In a $\Delta ABC$,the expression $\left( \frac{a^2}{\sin A} + \frac{b^2}{\sin B} + \frac{c^2}{\sin C} \right) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ simplifies to,where $\Delta$ is the area of the triangle.

In a $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{a^2-b^2}{a^2+b^2}$ is equal to