Tan 3-1-tan 3-1-3 then the general value of - is

The correct option is C nπ/3+7π/36 Applying componendo and dividendo nbsp; nbsp;tan 3θ nbsp; 1+tan 3θ nbsp; +1 /tan 3θ nbsp; 1 tan 3θ nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

tan 3θ-1/tan ...

Question

$\frac{t a n 3 θ - 1}{t a n 3 θ + 1} = \sqrt{3}$ then the general value of $θ$ is

\frac{n π}{3} - \frac{π}{12}

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n π + \frac{7 π}{12}

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\frac{n π}{3} + \frac{7 π}{36}

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n π + \frac{π}{12}

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Solution

The correct option is C $\frac{n π}{3} + \frac{7 π}{36}$
Applying componendo and dividendo
$\frac{tan 3 θ - 1 + tan 3 θ + 1}{tan 3 θ - 1 - tan 3 θ - 1} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$
$\Rightarrow - 2 tan 3 θ = \frac{4 + 2 \sqrt{3}}{2} \Rightarrow tan 3 θ = - 2 - \sqrt{3}$
$\Rightarrow 3 θ = n π + \frac{7 π}{12} \Rightarrow θ = \frac{n π}{3} + \frac{7 π}{36}$

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tan3θ−1tan3θ+1=√3 then the general value of θ is

The correct option is C nπ3+7π36Applying componendo and dividendo tan3θ−1+tan3θ+1tan3θ−1−tan3θ−1=√3+1√3−1⇒−2tan3θ=4+2√32⇒tan3θ=−2−√3⇒3θ=nπ+7π12⇒θ=nπ3+7π36

$\frac{t a n 3 θ - 1}{t a n 3 θ + 1} = \sqrt{3}$ then the general value of $θ$ is