First slide

Multiple and sub- multiple Angles

Question

$A = \frac{π}{7} B = \frac{2 π}{7} C = \frac{4 π}{7} and \cos A \cos B \cos C = \frac{- 1}{8} then$ $\sum \tan A . \tan B =$

Difficult

A

7
B

-7
C

$\frac{1}{7}$
D

1

Solution

$\begin{array}{l} W e h a v e A + B + C = π \\ N o w c os (A + B + C) = \cos (A + B) \cos C - \sin (A + B) \sin C \\ \Rightarrow \frac{\cos (A + B + C)}{\cos A \cos B \cos C} = 1 - \tan A \tan B - \tan B \tan C - \tan C \tan A \\ \Rightarrow \frac{\cos π}{- 1 / 8} = 1 - \sum \tan A \tan B \\ \Rightarrow 8 = 1 - \sum \tan A \tan B \\ \Rightarrow \sum \tan A \tan B = - 7 \end{array}$

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