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If x = a ${(\sec θ + \tan θ)}^{2}, y = b {(\sec θ - \tan θ)}^{2} t h e n x^{2} y^{2} =$

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$a b \sec θ$

$a^{2} b^{2} \sec θ$

$a^{2} b^{4}$

$a^{2} b^{2}$

answer is D.

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Detailed Solution

$x^{2} y^{2} = a^{2} {(\sec θ + \tan θ)}^{4} . b^{2} {(\sec θ - \tan θ)}^{4} = a^{2} b^{2} {(\sec^{2} θ - \tan^{2} θ)}^{2} = a^{2} b^{2}$

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