First slide

Multiple and sub- multiple Angles

Question

If $α, β, γ$ and $δ$ are the solution of the equation $\tan (θ + \frac{π}{4}) = 3 \tan 3 θ$ , no two of which have equal tangents then the value of
$\tan α + \tan β + \tan γ + \tan δ =$

Moderate

A

$\frac{1}{3}$
B

$\frac{8}{3}$
C

$- \frac{8}{3}$
D

0

Solution

$I f α, β, γ a n d δ$ are the solutions of the equation $\tan (θ + \frac{π}{4}) = 3 \tan 3 θ$
$\begin{array}{l} \Rightarrow \frac{1 + \tan θ}{1 - \tan θ} = 3 \times \frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ} \\ \Rightarrow \frac{1 + t}{1 - t} = 3 (\frac{3 t - t^{3}}{1 - 3 t^{2}}) (p u t t i n g t = \tan θ) \\ \Rightarrow 3 t^{4} - 6 t^{2} + 8 t - 1 = 0 \\ ∴ s u m o f r o o t s = t_{1} + t_{2} + t_{3} + t_{4} = 0 \\ \Rightarrow \tan α + \tan β + \tan γ + \tan δ = 0 \end{array}$

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