Question

In $\Delta$ABC, if $\cot \frac{A}{2}\cdot \cot \frac{B}{2}=c$, $\cot \frac{B}{2}\cdot \cot \frac{C}{2}=a$ and $\cot \frac{C}{2}\cdot \cot \frac{A}{2}=b$ then $\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}=$.

• A. $1$
• B. $2$
• C. abc
• D. $\Delta$

Answer 1

Answer: (B)

$2$

$\cot \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$

$\cot \frac{B}{2}=\sqrt{\frac{s(s-a)}{(s-c)(s-a)}}$

$\cot \frac{C}{2}=\sqrt{\frac{s(s-c)}{(s-b)(s-a)}}$

$\cot \frac{A}{2}\cot \frac{B}{2}.c$

$\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}.\sqrt{\frac{s(s-a)}{(s-c)(s-a)}}.c$

$\frac{s}{s-c}=c$

$\frac{1}{s-c}=\frac{c}{s}$

similarly

$\cot \frac{B}{2}\cot \frac{C}{2}=a$

$\frac{s}{s-a}=a$

$\frac{1}{s-a}=\frac{a}{s}$

$\cot \frac{C}{2}.\cot \frac{A}{2}=b$

Similarly

$\frac{1}{s-b}=\frac{b}{s}$

$\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}=\frac{a+b+c}{s}=\frac{a+b+c}{\frac{a+b+c}{2}}=2$

$s=\frac{a+b+c}{2}$