If tan-tan 2 -tan 3 -0 then -

The correct option is B tan θ + tan 2 θ + tan 3 θ =0 ⇒ tan θ +tan 2 θ + tan (θ + 2θ)=0 ⇒ tan θ + tan 2 θ + tan θ + tan 2 θ/1 tan θ. tan 2 θ=0 ⇒ (tan θ + tan 2 ...

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Byju's Answer

Standard XII

Mathematics

Differentiation of Inverse Trigonometric Functions

If tanθ+tan 2...

Question

If $t a n θ + t a n 2 θ + t a n 3 θ = 0 t h e n θ =$

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Solution

The correct option is A
$t a n θ + t a n 2 θ + t a n 3 θ = 0 \Rightarrow t a n θ + t a n 2 θ + t a n (θ + 2 θ) = 0 \Rightarrow t a n θ + t a n 2 θ + \frac{t a n θ + t a n 2 θ}{1 - t a n θ . t a n 2 θ} = 0$
$\Rightarrow (t a n θ + t a n 2 θ) (2 - t a n θ t a n 2 θ) = 0 \Rightarrow t a n θ + t a n 2 θ = 0 o r t a n θ . t a n 2 θ = 2$
$\Rightarrow t a n θ = t a n (- 2 θ) \Rightarrow θ = n π - 2 θ \Rightarrow θ = \frac{n π}{3}$
$t a n θ . t a n 2 θ = 2 \Rightarrow t a n θ . \frac{2 t a n θ}{1 - t a n^{2} θ} = 2 \Rightarrow t a n^{2} θ = 1 - t a n^{2} θ \Rightarrow t a n^{2} θ = \frac{1}{2} \Rightarrow θ = n π \pm T a n^{-} 1 (\frac{1}{\sqrt{2}}) .$

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If tanθ+tan2θ+tan3θ=0thenθ=

If $t a n θ + t a n 2 θ + t a n 3 θ = 0 t h e n θ =$