Question

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 $x+y+z=?$

Answer 1

We know that from Napier's analogies

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

$\therefore \tan \left[\frac{B-C}{2}\right]tan\frac{A}{2}=\left(\frac{b-c}{b+c}\right)=x\left(given\right).....\left(1\right)$

$similarly\Rightarrow y=\tan \left[\frac{C-A}{2}\right]\tan \frac{B}{2}=\left(\frac{c-a}{c+a}\right).....\left(2\right)$

$and$

$\Rightarrow z=\tan \left[\frac{A-B}{2}\right]\tan \frac{C}{2}=\left(\frac{a-b}{a+b}\right).....\left(3\right)$

$\therefore x+y+z=\left(\frac{b-c}{b+c}\right)+\left(\frac{c-a}{c+a}\right)+\left(\frac{a-b}{a+b}\right)$

$\Rightarrow \left[\frac{(b-c)\left(ac+bc+a^2+ab\right)+\left(c-a\right)\left(ab+ac+b^2+bc\right)+\left(a-b\right)\left(bc+ab+c^2+ac\right)}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\right]$

$\Rightarrow \left[\frac{0}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\right]\Rightarrow 0$