If the equation tan-tan 2 -tan-tan 2 -1 then - is equal to

The correct option is B tan θ + tan 2θ =1 tanθ tan 2θ tan θ+tan 2θ/1 tan θ tan 2θ=1 ex ⇒ex tan 3 θ =1 0ex ⇒ex 3 θ =n π +π 4 0ex ⇒ex θ =nπ 3 + π 12, n∈ I ...

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Standard XII

Mathematics

Domain

If the equati...

Question

If the equation $t a n θ + t a n 2 θ + t a n θ t a n 2 θ = 1 t h e n θ$ is equal to

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nπ+π/2, n ϵ I

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Solution

The correct option is B

$t a n θ + t a n 2 θ = 1 - t a n θ t a n 2 θ$
$\frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} = 1 \Rightarrow t a n 3 θ = 1$
$\Rightarrow 3 θ = n π + \frac{π}{4}$
$\Rightarrow θ = \frac{n π}{3} + \frac{π}{12}, n \in I$

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Similar questions

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If the equation tanθ+tan2θ+tanθtan2θ = 1 then θ is equal to

The correct option is B tanθ+tan2θ=1−tanθtan2θ tanθ+tan2θ1−tanθtan2θ=1⇒tan3θ=1 ⇒3θ=nπ+π4 ⇒θ=nπ3+π12, n∈I

If the equation $t a n θ + t a n 2 θ + t a n θ t a n 2 θ = 1 t h e n θ$ is equal to

The correct option is B

$t a n θ + t a n 2 θ = 1 - t a n θ t a n 2 θ$
$\frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} = 1 \Rightarrow t a n 3 θ = 1$
$\Rightarrow 3 θ = n π + \frac{π}{4}$
$\Rightarrow θ = \frac{n π}{3} + \frac{π}{12}, n \in I$