Question

If A + B + C = 180°, then $\frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C}=$

(a) tan A tan B tan C
(b) 0
(c) 1
(d) None of these

Answer 1

(c) 1
Using tan(180-A) = -tan A, we get:


$$\begin{gathered} C=\mathrm{\pi }-(A+B) \\ \mathrm{Now}, \\ \frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C} \\ =\frac{\tan A+\tan B+\tan \left[\pi -(A+B)\right]}{\tan A\tan B\tan \left[\pi -(A+B)\right]} \\ =\frac{\tan A+\tan B-\tan (A+B)}{-\tan A\mathit{}\tan B\mathit{}t\mathrm{an}(A+B)} \\ =\frac{\tan A+\tan B-{\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}}}{-\tan A\tan B\times {\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}}} \end{gathered}$$

$$\begin{gathered} =\frac{\tan A+\tan B-{\tan }^{2}A\tan B-\tan A{\tan }^{2}B-\tan A-\tan B}{-{\tan }^{2}A\mathit{}\tan B-\tan A{\tan }^{2}B} \\ =\frac{-{\tan }^{2}A\tan B-\mathrm{tanA}{\tan }^2\mathrm{B}}{-{\tan }^2\mathrm{AtanB}-\mathrm{tanA}{\tan }^2\mathrm{B}} \\ =1 \end{gathered}$$