$I f x = \sec θ - \tan θ, y = \cos e c θ + \cot θ, t h e n y =$

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$\frac{x}{1 + x}$

$\frac{1 + x}{1 - x}$

$\frac{x}{1 - x}$

$\frac{1 - x}{1 + x}$

answer is A.

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Detailed Solution

$\begin{array}{l} x = \sec θ - \tan θ = \frac{1 - \sin θ}{\cos θ}, y = \cos e c θ + \cot θ = \frac{1 + \cos θ}{\sin θ} \\ x y = (\frac{1 - \sin θ}{\cos θ}) (\frac{1 + \cos θ}{\sin θ}) = \frac{1 + \cos θ - \sin θ - \sin θ \cos θ}{\sin θ \cos θ} = \frac{1 + \cos θ - \sin θ}{\sin θ \cos θ} - 1 \\ \Rightarrow x y + 1 = \frac{1 + \cos θ - \sin θ}{\sin θ \cos θ} = \frac{(\cos^{2} θ + \sin^{2} θ) + \cos θ - \sin θ}{\sin θ \cos θ} \\ = \frac{\cos θ (1 + \cos θ) - \sin θ (1 - \sin θ)}{\sin θ \cos θ} = \frac{1 + \cos θ}{\sin θ} - \frac{1 - \sin θ}{\cos θ} = y - x \\ ∴ x y + 1 = y - x \Rightarrow y (1 - x) = 1 + x \Rightarrow y = \frac{1 + x}{1 - x} . \end{array}$