Question

In a triangle ABC, if $\cot A+\cot B+\cot C=\cot \theta$, then prove that $\frac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C}=\tan \theta$ and the possible value of $\theta$ is ${30}^o$.

Answer 1

$1+\cos A\cos B\cos C=\cos A\sin B\sin C+\cos B\sin C\sin A+\cos C\sin A\sin B$
$\therefore E=\frac{\sin A\sin B\sin C}{\cos A\sin B\sin C+\sin A\cos B\sin C+\sin A\sin B\cos C}$
Divide above and below by $\sin A\sin B\sin C$
$\therefore E=\frac{1}{cotA+\cot B+\cot C}=\frac{1}{\cot \theta }=\tan \theta$
$2$nd [art
${\cot }^2\theta =\sum {\cot }^{2}A+2\sum \cot B\cot C$
$=\frac{1}{2}[2\sum {\cot }^{2}A+4\sum \cot B\cot C]$
$=\frac{1}{2}[\sum (\cot A-\cot B{)}^2+2\sum \cot B\cot C+4\sum \cot B\cot C]$
$=\frac{1}{2}[6+\sum (\cot A-\cot B{)}^2]$
$\because \sum \cot B\cot C=1$
$=3+$something$+$ive
$\therefore {\cot }^2\theta \ge 3$ or $\tan \theta \le \frac{1}{\sqrt{3}}\therefore \theta ={30}^o$.