In any $\Delta ABC$,if $a \cos B = b \cos A$,then the triangle is

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 $\triangle ABC$,with usual notations,if $\cos \frac{B}{2} = \sqrt{\frac{c+a}{2a}}$,then $a^2 =$

Consider a triangle $ABC$ and let $a, b$ and $c$ denote the lengths of the sides opposite to vertices $A, B$ and $C$ respectively. Suppose $a=6, b=10$ and the area of the triangle is $15 \sqrt{3}$. If $\angle ACB$ is obtuse and if $r$ denotes the radius of the incircle of the triangle,then $r^2$ is equal to

In any $\triangle ABC$,$\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} =$

If the sides of a triangle are in $A.P.$,then the cotangents of its half-angles will be in