Mathematics

In $$\Delta ABC$$. If $$x=\tan\left(\dfrac{B-C}{2}\right)\tan\dfrac{A}{2}, y=\tan\left(\dfrac{C-A}{2}\right)\tan\dfrac{B}{2}, z=\tan\left(\dfrac{A-B}{2}\right)\tan\dfrac{C}{2}$$, then $$x+y+z$$ (in terms of $$x,y,z$$ only) is 


ANSWER

$$xyz$$


SOLUTION
In $$ \triangle ABC$$

$$\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$$

$$\implies x=\dfrac{b-c}{b+c}\implies\dfrac{-c}{b}$$

Similarly $$y=\dfrac{-a}{c},z=\dfrac{-b}{a}$$

These on adding gives $$\dfrac{-(ac^2+ba^2+cb^2)}{(abc)^2}$$
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