In a $\Delta ABC$,$\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2} = $ (in $/ a^2$)

In a triangle $ABC$,with usual notations,if $3b = a + c$,then $\cot \frac{A}{2} \cdot \cot \frac{C}{2} = $

In a $\triangle ABC$,$\angle C = 60^{\circ}$ and $\angle A = 75^{\circ}$. If $D$ is a point on $AC$ such that the area of $\triangle BAD$ is $\sqrt{3}$ times the area of $\triangle BCD$,then the measure of $\angle ABD$ is (in $^{\circ}$)

In $\triangle ABC$,if $\theta$ is any angle,then $b \cos (C+\theta) + c \cos (B-\theta) =$

In $\Delta ABC,\, \left( {\cot \frac{A}{2} + \cot \frac{B}{2}} \right)\,\left( {a{{\sin }^2}\frac{B}{2} + b{{\sin }^2}\frac{A}{2}} \right) =$

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