Question

If $A,B,C$ are the angles of a triangle and if none of them is equal to $\frac{\pi }{2}$, then find
$\tan A+\tan B+\tan C-\tan A\tan B\tan C$$+$ $\cot A\cot B+\cot B\cot C+\cot C\cot A$

Answer 1

Given

In $\Delta ABC$

$\tan (A+B+C)=\pi$

$\tan (A+B+C)=\tan \pi$

$\frac{(\tan A+\tan B+\tan C-\tan A\tan B\tan C)}{(1-\tan A\tan B-\tan B\tan C-\tan A\tan C)}=0$

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

$⟹\frac{(\tan A+\tan B+\tan C)}{\left(\tan A\tan B\tan C\right)}=1$

$⟹\cot A\cot B+\cot B\cot C+\cot C\cot A=1$

So, $\tan A+\tan B+\tan C-\tan A\tan B\tan C+\cot A\cot B+\cot B\cot C+\cot C\cot A$

$⟹0+1=1$