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If $θ_{1}$ and $θ_{2}$ are the angles made by the lines $(x^{2} + y^{2}) (\cos^{2} θ \sin^{2} α + \sin^{2} θ) = (x \tan α - y \sin θ)^{2}$ with positive axis of $X$ in the anticlock wise direction, if $θ = \frac{π}{6}, t a n θ_{1} + t a n θ_{2}$ equals

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$(- \frac{8}{3}) \sin^{2} α$

$(- 8 \sqrt{3}) cosec 2 α$

$(- \frac{8}{3}) cosec 2 α$

$(- 4 \sqrt{3}) cosec 2 α$

answer is C.

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Detailed Solution

$x^{2} (\cos^{2} θ \sin^{2} α + \sin^{2} θ) + y^{2} (\cos^{2} θ \sin^{2} α + \sin^{2} θ)$

$= x^{2} \tan^{2} α + y^{2} \sin^{2} θ - 2 x y \tan α \sin θ$

$Now, \tan θ_{1} + \tan θ_{2}$

$= m_{1} + m_{2} = \frac{- 2 h}{b} = \frac{- (- 2 \tan α \sin θ)}{- \cos^{2} θ \sin^{2} α} at θ = \frac{π}{6}$

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