If $x = a \sec θ + b \tan θ, y = b \sec θ + a \tan θ then x^{2} - y^{2} =$

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$4 absec θ \cdot \tan θ$

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

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

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

answer is B.

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Detailed Solution

Given $x = asec θ + \tan θ$
$\begin{array}{l} (x)^{2} = (asec θ + btan θ)^{2} \\ x^{2} = a^{2} \sec^{2} θ + b^{2} \tan^{2} θ + 2 absec θtan θ \end{array}$
And $y = bsec θ + atan θ$
S.O.B.S
$\begin{array}{l} y^{2} = (bsec θ + atan θ)^{2} \\ y^{2} = b^{2} \sec^{2} θ + a^{2} \tan^{2} θ + 2 absec θtan θ \\ ∴ x^{2} - y^{2} = (a^{2} \sec^{2} θ + b^{2} \tan^{2} θ + 2 absec θtan θ) \\ - (b^{2} \sec^{2} θ + a^{2} \tan^{2} θ + 2 absec θtan θ) \begin{array}{l} x^{2} - y^{2} = a^{2} \sec^{2} θ + b^{2} \tan^{2} θ + 2 absec θtan θ - b^{2} \sec^{2} θ - a^{2} \tan^{2} θ - 2 absec θtan θ \\ x^{2} - y^{2} = a^{2} (\sec^{2} θ - \tan^{2} θ) + b^{2} (\tan^{2} θ - \sec^{2} θ) x^{2} - y^{2} = a^{2} (1) + b^{2} (- 1) x^{2} - y^{2} = a^{2} - b^{2} (∵ \sec^{2} θ - \tan^{2} θ = 1) \end{array} \end{array}$