Question

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

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

The correct option is D 2

$\tan A/2=\frac{\Delta }{(s-a)s}$

$s\frac{\tan A/2}{\Delta }=\frac{1}{(s-a)}$

$\frac{S}{\Delta }(\tan A/2+\tan B/2+\tan C/2)$

$\cot A/2=\frac{(s-a)s}{\Delta },\cot B/2=\frac{(s-b)s}{\Delta },\cot C/2=\frac{(s-c)s}{\Delta }$

$\frac{(s-c)}{(s-c)}\times \frac{(s-a)s}{\Delta }\times \frac{(s-b)s}{\Delta }=C$

$\frac{s}{(s-c)}=C$

$\frac{1}{(s-c)}=\frac{c}{s}$, ---(i) Similarly$\frac{1}{(s-b)}=\frac{b}{s},...\left(ii\right)\frac{1}{(s-a)}=\frac{a}{s}$....(iii)

after observing eq(i),eq(ii) and eq(iii)

So the equation asked in the qs is equal to $\frac{c}{s}+\frac{b}{s}+\frac{a}{s}$

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

$=2$.