Question

In a triangle $ABC$ $\frac{1}{(a-b)(a-c)}\tan \frac{A}{2}+\frac{1}{(b-c)(b-a)}\tan \frac{B}{2}+\frac{1}{(c-a)(c-b)}\tan \frac{C}{2}=$

• A. $\frac{1}{\Delta }$
• B. $\frac{2}{\Delta }$
• C. $\frac{3}{\Delta }$
• D. $\frac{4}{\Delta }$

Answer 1

The correct option is A $\frac{1}{\Delta }$

We have,

$\frac{1}{\left(a-b\right)\left(a-c\right)}\tan \frac{A}{2}+\frac{1}{\left(b-c\right)\left(b-a\right)}\tan \frac{B}{2}+\frac{1}{\left(c-a\right)\left(c-b\right)}\tan \frac{C}{2}$

On multiplying and dividing

$S\left(s-a\right)\left(a-b\right)\left(a-c\right)S\left(s-b\right)\left(b-c\right)\left(b-a\right)S\left(s-c\right)\left(c-a\right)\left(c-b\right)$

$=\frac{1}{\Delta }\left[\frac{\left(s-b\right)\left(s-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(s-c\right)\left(s-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(s-a\right)\left(s-c\right)}{\left(c-a\right)\left(c-b\right)}\right]=\frac{1}{\Delta }$

Hence, this is the answer.