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If $3 T a n^{4} α - 10 T a n^{2} α + 3 = 0$ then principal values of ‘ $α$ ’ are

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\pm 45^{0}, \pm 36^{0}

\pm 30^{0}, \pm 60^{0}

\pm 75^{0}, \pm 36^{0}

\pm 60^{0}, \pm 15^{0}

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detailed solution

Correct option is B

$3 \tan^{4} α - 10 \tan^{2} α + 3 = 0$

$\Rightarrow 3 \tan^{4} α - 9 \tan^{2} α - \tan^{2} α + 3 = 0$

$\Rightarrow (3 \tan^{2} α - 1) (\tan^{2} α - 3) = 0$

$\Rightarrow \tan^{2} α = \frac{1}{3} = {(\frac{1}{\sqrt{3}})}^{2}$ $; \tan^{2} α = 3 = {(\sqrt{3})}^{2}$

$∴ α = \pm 30^{0}, \pm 60^{0}$

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If 3Tan4α−10Tan2α+3=0 then principal values of ‘α’ are