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if sec theta -tan theta=1/3,the value of [sec theta+tan theta] is

$\sec^{2} θ - \tan^{2} θ = 1 (\sec θ + \tan θ) (\sec θ - \tan θ) = 1 \sec θ + \tan θ = \frac{1}{\sec θ - \tan θ} = \frac{1}{(\frac{1}{3})} = 3$

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