Question

If $ABC$ is a triangle and $A+B+C={180}^{\circ },$ then $\tan A+\tan B+\tan C$ is

• A. $\tan A\tan B\tan C$
• B. $\frac{\tan A+\tan B}{\tan C}$
• C. $\tan \frac{A+B}{2}+\tan \frac{B+C}{2}$
• D. $cotA\cot B\cot C$

Answer 1

The correct option is A $\tan A\tan B\tan C$

$A+B+C={180}^o$

$A+B={180}^o-C$

$\tan (A+B)=\tan ({180}^o-C)$

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

$\frac{\tan A+\tan B}{1-\tan A\cdot \tan B}=-\tan C$

$\tan A+\tan B=\tan C(\tan A\cdot \tan B-1)$

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

Hence $\tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C$.