Simplify $\sqrt { \dfrac{ \mathrm{cosec} \theta - 1 } { \mathrm{cosec} \theta + 1 } } + \sqrt { \dfrac{ \mathrm{cosec} \theta + 1 } { \mathrm{cosec} \theta - 1 } }$

Answer:

$2sec\theta$

Step by Step Explanation:
1. \begin{align} & \sqrt { \dfrac{ \mathrm{cosec} \theta - 1 } { \mathrm{cosec} \theta + 1 } } + \sqrt { \dfrac{ \mathrm{cosec} \theta + 1 } { \mathrm{cosec} \theta - 1 } } \\ = & \sqrt { \dfrac{ ( \mathrm{cosec} \theta - 1 ) ( \mathrm{cosec} \theta - 1 ) } { ( \mathrm{cosec} \theta + 1 )( \mathrm{cosec} \theta - 1 ) } } + \sqrt { \dfrac{ ( \mathrm{cosec} \theta + 1 ) ( \mathrm{cosec} \theta + 1 ) } { ( \mathrm{cosec} \theta - 1 )( \mathrm{cosec} \theta + 1 ) } } \\ = & \sqrt { \dfrac{ ( \mathrm{cosec} \theta - 1 )^2 } { \cot^2 \theta } } + \sqrt { \dfrac{ ( \mathrm{cosec} \theta + 1 )^2 } { \cot^2 \theta } } \\ = & \dfrac{ 1 + \mathrm{cosec} \theta - 1 + \mathrm{cosec} \theta } { \cot \theta } \\ = & \dfrac{ 2\mathrm{cosec}\theta } { \cot \theta } \\ = & 2 \sec \theta \\ \end{align}

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