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

Question 114

Multiple Choice

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 } } }


A) 8tan3θ8 \tan ^ { 3 } \theta

B) 18sin3θ\frac { 1 } { 8 \sin ^ { 3 } \theta }

C) 1sin3θ\frac { 1 } { \sin ^ { 3 } \theta }

D) 14sin3θ\frac { 1 } { 4 \sin ^ { 3 } \theta }

E)  none \text { none }

Correct Answer:

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