The angle bisectors $BD$ and $CE$ of a $\triangle ABC$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively. Then,the ratio in which $I$ divides the angle bisector through $A$ is

In a $\triangle ABC$,if $A-B=120^{\circ}$ and $R=8r$,then find the value of $\frac{1+\cos C}{1-\cos C}$.

In $\triangle ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$ and $C = \frac{\pi}{2}$,then $A : B =$

In $\triangle ABC$,if $a=2$,$B=\tan ^{-1} \frac{1}{2}$ and $C=\tan ^{-1} \frac{1}{3}$,then $(A, b)$ equals

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:
$(A) b+c=4a$
$(B) b+c=2a$
$(C) \text{locus of point } A \text{ is an ellipse}$
$(D) \text{locus of point } A \text{ is a pair of straight lines}$

In $\triangle ABC$,if $x=\tan \left(\frac{B-C}{2}\right) \tan \frac{A}{2}$,$y=\tan \left(\frac{C-A}{2}\right) \tan \frac{B}{2}$,and $z=\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}$,then $(x+y+z)$ is equal to