If cosθ=−32 and sinα=−35 , where θ does not lie in the third quadrant, α∈Q3 then 2tanα+3tanθcot2θ+cosα is equal to

θ=150° sinα=−35 cosα=−45 tanα=34 2tanα+3tanθcot2θ+cosα=2(34)+3(−13)(3)2−45 =32−13−45=12115 =522

If cosθ=−32 and sinα=−35 , where θ does not lie in the third quadrant, α∈Q3 then 2tanα+3tanθcot2θ+cosα is equal to

θ=150° sinα=−35 cosα=−45 tanα=34 2tanα+3tanθcot2θ+cosα=2(34)+3(−13)(3)2−45 =32−13−45=12115 =522

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If $\cos θ = - \frac{\sqrt{3}}{2}$ and $\sin α = - \frac{3}{5}$ , where $θ$ does not lie in the third quadrant, $α \in Q_{3}$ then $\frac{2 \tan α + \sqrt{3} \tan θ}{\cot^{2} θ + \cos α}$ is equal to

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\frac{7}{22}

\frac{5}{22}

\frac{9}{22}

\frac{22}{5}

answer is B.

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Detailed Solution

θ = 150 °

$\sin α = - \frac{3}{5}$

$\cos α = - \frac{4}{5}$

$\tan α = \frac{3}{4}$

$\frac{2 \tan α + \sqrt{3} \tan θ}{\cot^{2} θ + \cos α} = \frac{2 (\frac{3}{4}) + \sqrt{3} (- \frac{1}{\sqrt{3}})}{{(\sqrt{3})}^{2} - \frac{4}{5}}$

$= \frac{\frac{3}{2} - 1}{3 - \frac{4}{5}} = \frac{\frac{1}{2}}{\frac{11}{5}}$

$= \frac{5}{22}$

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If cosθ=−32 and sinα=−35 , where θ does not lie in the third quadrant, α∈Q3 then 2tanα+3tanθcot2θ+cosα is equal to

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If $\cos θ = - \frac{\sqrt{3}}{2}$ and $\sin α = - \frac{3}{5}$ , where $θ$ does not lie in the third quadrant, $α \in Q_{3}$ then $\frac{2 \tan α + \sqrt{3} \tan θ}{\cot^{2} θ + \cos α}$ is equal to