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

In $\triangle ABC$, with usual notation, if $\cot A, \cot B, \cot C$ are in arithmetic progression, then

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

If $b+c=3a$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

The sides of a triangle are $\sin \alpha$,$\cos \alpha$,and $\sqrt{1 + \sin \alpha \cos \alpha}$ for some $0 < \alpha < \frac{\pi}{2}$. Then the greatest angle of the triangle is.....$^o$

In a right-angled $\triangle ABC$, the measures of the angles are in an Arithmetic Progression ($A$.$P$.). If its smallest side is $4 \text{ units}$, then the area of $\triangle ABC$ is: