Question

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

• A. $xyz$
• B. $2xyz$
• C. $-xyz$
• D. $\frac{1}{2}xyz$

Answer 1

The correct option is A $xyz$

In $△ABC$

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

$⟹x=\frac{b-c}{b+c}⟹\frac{-c}{b}$

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

These on adding gives $\frac{-(a{c}^2+b{a}^2+c{b}^2)}{(abc{)}^2}$