If 1-tan 2 - 2 - - 1 2- then the general value of - is

The correct option is B nπ +π 6 , n∈ Z 1 tan^2θ^2θ=12 ⇒cos^2θ tan^2θcos^2θ=12 =cos^2θ sin^2θ=12 cos2θ=12 2θ=π3+2nπ θ=π6+nπ ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

If 1-tan 2 ...

Question

If $\frac{1 - {tan}^{2} θ}{{sec}^{2} θ} = \frac{1}{2}$ , then the general value of $θ$ is

n π + \frac{π}{6}, n \in Z

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n π - \frac{π}{6}, n \in Z

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2 n π + \frac{π}{6}, n \in Z

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none of these

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Solution

The correct option is B $n π + \frac{π}{6}, n \in Z$
$\frac{1 - t a n^{2} θ}{{sec}^{2} θ} = \frac{1}{2} \Rightarrow {cos}^{2} θ - t a n^{2} θ {cos}^{2} θ = \frac{1}{2} = {cos}^{2} θ - {sin}^{2} θ = \frac{1}{2} cos 2 θ = \frac{1}{2} 2 θ = \frac{π}{3} + 2 n π θ = \frac{π}{6} + n π$

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Similar questions

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If 1−tan2θsec2θ=12, then the general value of θ is

The correct option is B nπ+π6,n∈Z1−tan2θsec2θ=12⇒cos2θ−tan2θcos2θ=12=cos2θ−sin2θ=12cos2θ=122θ=π3+2nπθ=π6+nπ

If $\frac{1 - {tan}^{2} θ}{{sec}^{2} θ} = \frac{1}{2}$ , then the general value of $θ$ is

The correct option is B $n π + \frac{π}{6}, n \in Z$
$\frac{1 - t a n^{2} θ}{{sec}^{2} θ} = \frac{1}{2} \Rightarrow {cos}^{2} θ - t a n^{2} θ {cos}^{2} θ = \frac{1}{2} = {cos}^{2} θ - {sin}^{2} θ = \frac{1}{2} cos 2 θ = \frac{1}{2} 2 θ = \frac{π}{3} + 2 n π θ = \frac{π}{6} + n π$