Question

If $A+B+C={180}^{\circ },$ then $\frac{tanA+tanB+tanC}{tanAtanBtanC}$ is equal to

• A. tanA tanB tanC
  
• B. 0
  
• C. 1
  
• D. none of these

Answer 1

The correct option is C 1


Using $tan(180-A)=-tanA,$ we get:
$C=\pi -(A+B)$
Now,
$\frac{tanA+tanB+tanC}{tanAtanBtanC}$
$=\frac{tanA+tanB+tan[\pi -(A+B\left)\right]}{tanAtanBtan[\pi -(A+B\left)\right]}$
$=\frac{tanA+tanB-tan(A+B)}{-tanAtanBtan(A+B)}$
$=\frac{tanA+tanB-\frac{tanA+tanB}{1-tanAtanB}}{-tanAtanB\times \frac{tanA+tanB}{1-tanAtanB}}$
$=\frac{tanA+tanB-ta{n}^{2}AtanB-tanAta{n}^{2}B-tanA-tanB}{-ta{n}^{2}AtanB-tanAta{n}^{2}B}$
$=\frac{-ta{n}^{2}AtanB-tanAta{n}^{2}B}{-ta{n}^{2}AtanB-tanAta{n}^{2}B}$
=1