First slide

Multiple and sub- multiple Angles

Question

For $0 < θ < \frac{π}{2}, \tan θ + \tan 2 θ + \tan 3 θ = 0$ if

Moderate

A

$\tan θ = 0$
B

$\tan 2 θ = 0$
C

$\tan 3 θ = 2$
D

$\tan θ . \tan 2 θ = 2$

Solution

Clearly, $\tan θ \neq 0 a n d \tan 2 θ \neq 0 f o r 0 < θ < \frac{π}{2}$
$W e h a v e$ $\tan 3 θ = \tan (2 θ + θ)$
$\begin{array}{l} \Rightarrow \tan 3 θ = \frac{\tan 2 θ + \tan θ}{1 - \tan 2 θ \tan θ} \\ N o w 0 = \tan θ + \tan 2 θ + \tan 3 θ \\ = \tan 3 θ (1 - \tan 2 θ \tan θ) + \tan 3 θ \\ = \tan 3 θ (2 - \tan 2 θ \tan θ) \\ \Rightarrow \tan 3 θ = 0 (o r) \tan 2 θ \tan θ = 2 \end{array}$

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