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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

ConceptTrigonometric Functions of Sum and Difference of Two Angles

#### 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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