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$$\frac{1}{\sec\theta\ -\tan\theta\ }=\frac{\sec\theta\ +\tan\theta\ }{\sec^2\theta\ -\tan^2\theta\ }=\left(\sec\theta\ +\tan\theta\ \right).$$
$$\ \left(we\ know\ \sec^2\theta\ -\tan^2\theta\ =1\right)$$
A is correct choice.
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