With usual notations in $\triangle ABC$, if $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = \frac{3a}{2}$, then:

In a triangle $ABC$ with a fixed base $BC$,the vertex $A$ moves such that $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$. If $a, b,$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B,$ and $C$,respectively,then:
$(A) b+c=4a$
$(B) b+c=2a$
$(C) \text{locus of point } A \text{ is an ellipse}$
$(D) \text{locus of point } A \text{ is a pair of straight lines}$

If $A, B, C$ are the angles of a triangle,then $\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = $

$A$ tower is situated on a horizontal plane. Two points lie on the line passing through the base of the tower,at distances $a$ and $b$ from the base. The angles of elevation of the top of the tower from these points are $\alpha$ and $90^\circ - \alpha$. If the line segment joining the two points subtends an angle $\theta$ at the top of the tower,find the height of the tower.

In a $\triangle ABC$, $a, b, c$ are the sides of the triangle opposite to the angles $A, B, C$ respectively. Then, the value of $a^{3} \sin (B-C) + b^{3} \sin (C-A) + c^{3} \sin (A-B)$ is equal to

In an isosceles triangle $ABC$,$\angle C = \angle A$. If the point of intersection of the bisectors of internal angles $\angle A$ and $\angle C$ divides the median of side $AC$ in the ratio $3 : 1$ (from vertex $B$ to side $AC$),then the value of $\csc \frac{B}{2}$ is equal to