Question
If in a $$\vartriangle ABC,3a = b + c$$ then $$\tan \frac{B}{2} \cdot \tan \frac{C}{2}$$ is equal to
A.
$$\tan \frac{A}{2}$$
B.
$$1$$
C.
$$2$$
D.
None of these
Answer :
None of these
Solution :
$$\eqalign{
& 3\sin \,A = \sin \,B + \sin \,C \cr
& \Rightarrow \,\,6\sin \frac{A}{2} \cdot \cos \frac{A}{2} \cdot = 2\sin \frac{{B + C}}{2} \cdot \cos \frac{{B - C}}{2} \cr
& \Rightarrow \,\,3\cos \frac{{B + C}}{2} = \cos \frac{{B - C}}{2} \cr
& \Rightarrow \,\,3\left( {\cos \frac{B}{2} \cdot \cos \frac{C}{2} - \sin \frac{B}{2} \cdot \sin \frac{C}{2}} \right) = \cos \frac{B}{2} \cdot \cos \frac{C}{2} + \sin \frac{B}{2}\sin \frac{C}{2} \cr
& \Rightarrow \,\,2\cos \frac{B}{2} \cdot \cos \frac{C}{2} = 4\sin \frac{B}{2} \cdot \sin \frac{C}{2} \cr
& \Rightarrow \,\,\tan \frac{B}{2} \cdot \tan \frac{C}{2} = \frac{1}{2}. \cr} $$