If $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$,then $A, B, C$ are in

Let $ABC$ be a triangle such that $AB=4, BC=5$ and $CA=6$. Choose points $D, E, F$ on $AB, BC, CA$ respectively,such that $AD=2, BE=2, CF=2$. Then find the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$.

If $R : r_1 : r = 5 : 12 : 2$, then $r + r_3 + r_2 - r_1 = $

If $b = 3, c = 4$ and $B = \frac{\pi}{3}$,then the number of triangles that can be constructed is

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

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.