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Let x=2tanθx = 2 \tan \theta 0θ<π20 \leq \theta < \frac { \pi } { 2 }

Question 51

Multiple Choice

Let x=2tanθx = 2 \tan \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) cos3θ8sin2θ\frac { \cos ^ { 3 } \theta } { 8 \sin ^ { 2 } \theta }

C) sec3θ\sec ^ { 3 } \theta

D) cos2θ4sinθ\frac { \cos ^ { 2 } \theta } { 4 \sin \theta }

E) 8cosθsin2θ\frac { 8 \cos \theta } { \sin ^ { 2 } \theta }

Correct Answer:

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