Question:
If $A+B+C=\pi$, then $\frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}$ is equal to
(a) tan A tan B tan C
(b) 0
(c) 1
(d) None of these
Solution:
(c) 1
π = 180°
Using tan(180 – A) = -tan A, we get:
$C=\pi-(A+B)$
Now,
$\frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}$
$=\frac{\tan A+\tan B+\tan [\pi-(A+B)]}{\tan A \tan B \tan [\pi-(A+B)]}$
$=\frac{\tan A+\tan B-\tan (A+B)}{-\tan A \tan B t \operatorname{an}(A+B)}$
$=\frac{\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A \tan B}}{-\tan A \tan B \times \frac{\tan A+\tan B}{1-\tan A \tan B}}$
$=\frac{\tan A+\tan B-\tan ^{2} A \tan B-\tan A \tan ^{2} B-\tan A-\tan B}{-\tan ^{2} A \tan B-\tan A \tan ^{2} B}$
$=\frac{-\tan ^{2} A \tan B-\tan A \tan ^{2} B}{-\tan ^{2} A \tan B-\tan A \tan ^{2} B}$
$=1$