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If $x = a {(secθ + tanθ)}^{2}$ and $y = b {(secθ - tanθ)}^{4},$ then $x^{4} y^{2} =$

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$ab secθ$

$a^{2} b^{2} tanθ$

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

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

answer is D.

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Detailed Solution

$x = a {(secθ + tanθ)}^{2}$

$y = b {(secθ - tanθ)}^{4},$

$\begin{array}{rcl} \Rightarrow & y = \frac{b}{{(secθ + tanθ)}^{4}} \\ y & = & \frac{b}{\frac{x^{2}}{a^{2}}} \\ x^{2} y & = & a^{2} b \\ x^{4} y^{2} & = & a^{4} b^{2} \end{array}$

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