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Let x=2sinθx = 2 \sin \theta 0θ<π20 \leq \theta < \frac { \pi } { 2 }

question 114

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Let x=2sinθx = 2 \sin \theta , 0θ<π20 \leq \theta < \frac { \pi } { 2 } ) Simplify the expression. 1x24+x2\frac { 1 } { x ^ { 2 } \sqrt { 4 + x ^ { 2 } } }


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