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Question
If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to
Options
tan A tan B tan C
0
1
None of these
Solution
1
π = 180°
Using tan(180 – A) = -tan A, we get:
\[C = \pi - (A + B)\]
Now,
\[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\]
\[ = \frac{\tan A + \tan B + \tan\left[ \pi - (A + B) \right]}{\tan A \tan B \tan\left[ \pi - (A + B) \right]}\]
\[ = \frac{\tan A + \tan B - \tan(A + B)}{- \tan A \tan B tan(A + B)}\]
\[ = \frac{\tan A + \tan B - \frac{\tan A + \tan B}{1 - \tan A \tan B}}{- \tan A \tan B \times \frac{\tan A + \tan B}{1 - \tan A \tan B}}\]
\[ = \frac{- \tan^2 A\tan B - tanA \tan^2 B}{- \tan^2 A \tan B - \tan A \tan^2 B}\]
\[ = 1\]
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