Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

Then $\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}=?$

• A. $\frac{4s^2}{a^2+b^2+c^2}$
• B. $\frac{a^2+b^2+c^2}{2s}$
• C. $\frac{a^2+b^2+c^2}{3s}$
• D. None of these

Answer 1

The correct option is A $\frac{4s^2}{a^2+b^2+c^2}$

Formula used:

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

$\tan B/2=\frac{s(s-b)}{\Delta };\cot C/2=\frac{s}{(s-c)}\Delta$

2) $\Delta =\frac{1}{2}bc\sin A⟹\sin A=\frac{2\Delta }{bc}$

Similarly:

$\sin B=\frac{2\Delta }{ca};\sin C=\frac{2\Delta }{ab}$

3) $a+b+c=2s$

Now; We have,

$\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}=\frac{\left[\frac{s(s-a)}{\Delta }\frac{s(s-b)}{\Delta }\frac{s(s-c)}{\Delta }\right]}{(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C})}$

$=\frac{\left(\frac{s(s-a)+s(s-b)+s(s-c)}{\Delta }\right)}{\frac{(b^2+c^2-a^2)\times bc}{2bc\times 2\Delta }+\frac{(a^2+c^2-b^2)\times ac}{2ac\times 2\Delta }+\frac{(a^2+b^2-c^2)\times ab}{2ab\times 2\Delta }}$

$=\frac{4s[3s-(a+b+c)]}{a^2+b^2+c^2}$

$=\frac{4s(3s-2s)}{a^2+b^2+c^2}=\frac{4s^2}{a^2+b^2+c^2}$

Hence;

$\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}=\frac{4s^2}{a^2+b^2+c^2}$