\begin{array} { l } { \cdot \text { In a } \Delta A B C \text { , prov

1.	\begin{array} { l } { \cdot \text { In a } \Delta A B C \text { , prove that } } \\ { \tan A + \tan B + \tan C = \tan A \tan B \tan C } \end{array}
Answer» Solution :- For any triangle the sum of the angles is equal to 180 degrees. If we take a triangle ABC, A + B + C = 180 degrees. A + B + C = 180 or A + B = 180 - C tan (A + B) = tan (180 - C) = tan C => (tan A + tan B)/(1 - tan Atan B) = tan C => tan A + tan B = tan C - tan Atan Btan C => tan A + tan B + tan C = tan Atan B*tan C

\begin{array} { l } { \cdot \text { In a } \Delta A B C \text { , prove that } } \\ { \tan A + \tan B + \tan C = \tan A \tan B \tan C } \end{array}

Answer»

Solution :- For any triangle the sum of the angles is equal to 180 degrees. If we take a triangle ABC, A + B + C = 180 degrees.

A + B + C = 180 or A + B = 180 - C

tan (A + B) = tan (180 - C) = tan C

=> (tan A + tan B)/(1 - tan A*tan B) = tan C

=> tan A + tan B = tan C - tan A*tan B*tan C

=> tan A + tan B + tan C = tan A*tan B*tan C