The correct option is D 3nπ2+π2 Given tan(23θ)=√(3) tan2θ3=tanπ3 2θ3=nπ+π3 θ=3 nπ2+π2 ...

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General Solution of Sin Theta = Sin Alpha

tan23θ=√3, th...

Question

$tan (\frac{2}{3} θ) = \sqrt{3}$ , then $θ$ is

\frac{n π}{2} + \frac{π}{6}

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n π

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\frac{3 n π}{2} - \frac{π}{4}

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\frac{3 n π}{2} + \frac{π}{2}

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Solution

The correct option is D $\frac{3 n π}{2} + \frac{π}{2}$
Given $tan (\frac{2}{3} θ) = \sqrt{3}$

$⟹ tan \frac{2 θ}{3} = tan \frac{π}{3}$

$⟹ \frac{2 θ}{3} = n π + \frac{π}{3}$

$⟹ θ = \frac{3 n π}{2} + \frac{π}{2}$

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tan(23θ)=√3, then θ is

The correct option is D 3nπ2+π2Given tan(23θ)=√3⟹tan2θ3=tanπ3⟹2θ3=nπ+π3⟹θ=3nπ2+π2

$tan (\frac{2}{3} θ) = \sqrt{3}$ , then $θ$ is

The correct option is D $\frac{3 n π}{2} + \frac{π}{2}$
Given $tan (\frac{2}{3} θ) = \sqrt{3}$

$⟹ tan \frac{2 θ}{3} = tan \frac{π}{3}$

$⟹ \frac{2 θ}{3} = n π + \frac{π}{3}$

$⟹ θ = \frac{3 n π}{2} + \frac{π}{2}$