Question

If $A,B,C$ are angles of triangle then
(i) $\tan A+\tan B+\tan C=\tan A\tan B\tan C$
(ii) $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$

Answer 1

$\to \left(i\right)\tan A+\tan B++\tan C$

$=\frac{\sin A\cos B+\sin B+\cos A}{\cos A\cos B}+\tan C$

$\frac{\sin (A+B)}{\cos B\cos B}+\frac{\sin C}{\cos C}$

$=\frac{\sin C}{\cos B\cos A}+\frac{\sin C}{\cos C}(\sin A-C=\sin C)$

$=\sin C\left(\frac{\cos C+\cos A\cos B}{\cos A\cos B\cos C}\right)$

$=\sin C\left(\frac{\cos (\pi -(A+B\left)\right)+\cos A\cos B}{\cos A\cos B\cos C}\right)$

$=\sin C\left(\frac{\cos A\cos B-\cos (A+B)}{\cos A\cos B\cos C}\right)$

$=\sin C\left(\frac{\cos A\cos B-\cos A\cos B+\sin A\sin B}{\cos A\cos B\cos C}\right)$

$=\frac{\sin A\sin B\sin C}{\cos A\cos B\cos C}=\tan A\tan B\tan C$

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

$\cot B\cos C=\frac{\tan A}{\sum \tan A},\cos A\cos C=\frac{\tan B}{\sum \tan A}$

$\Rightarrow Sum=\frac{\tan A+\tan B+\tan C}{\tan A+\tan B+\tan C}=1$