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If a + B + C = π, Then \[\Frac{\Tan a + \Tan B + \Tan C}{\Tan a \Tan B \Tan C}\] is Equal to - CBSE (Science) Class 11 - Mathematics

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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

 
  • 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,
\[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\]

\[ = \frac{\tan A + \tan B + \tan\left[ \pi - (A + B) \right]}{\tan A \tan B \tan\left[ \pi - (A + B) \right]}\]

\[ = \frac{\tan A + \tan B - \tan(A + B)}{- \tan A \tan B tan(A + B)}\]

\[ = \frac{\tan A + \tan B - \frac{\tan A + \tan B}{1 - \tan A \tan B}}{- \tan A \tan B \times \frac{\tan A + \tan B}{1 - \tan A \tan B}}\]

\[= \frac{\tan A + \tan B - \tan^2 A\tan B - \tan A \tan^2 B - \tan A - \tan B}{- \tan^2 A \tan B - \tan A \tan^2 B}\]
\[ = \frac{- \tan^2 A\tan B - tanA \tan^2 B}{- \tan^2 A \tan B - \tan A \tan^2 B}\]
\[ = 1\]
  Is there an error in this question or solution?

APPEARS IN

 RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 7: Values of Trigonometric function at sum or difference of angles
Q: 8 | Page no. 27
Solution If a + B + C = π, Then \[\Frac{\Tan a + \Tan B + \Tan C}{\Tan a \Tan B \Tan C}\] is Equal to Concept: Trigonometric Functions of Sum and Difference of Two Angles.
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