Let $x = 2 \tan \theta$ , $0 \leq \theta

Question

Let $x = 2 \tan \theta$ , $0 \leq \theta < \pi / 2$) Simplify the expression. $\frac { 1 } { x ^ { 2 } \sqrt { 4 + x ^ { 2 } } }$
A) $8 \tan ^ { 3 } \theta$

B) $\frac { \cos ^ { 3 } \theta } { 8 \sin ^ { 2 } \theta }$

C) $\sec ^ { 3 } \theta$

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

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

Quizplus · Accepted Answer

Given $x = 2 \tan \theta$, we substitute $x$ into the expression to get $\frac{1}{(2\tan\theta)^2\sqrt{4+(2\tan\theta)^2}}$. Simplifying inside the square root gives $4+4\tan^2\theta = 4(1+\tan^2\theta)$, which is $4\sec^2\theta$. Thus, the expression becomes $\frac{1}{4\tan^2\theta\cdot 2\sec\theta} = \frac{1}{8\sin^2\theta/\cos^2\theta \cdot 1/\cos\theta} = \frac{\cos^3\theta}{8\sin^2\theta}$.

Let x=2tan⁡θx = 2 \tan \thetax=2tanθ 0≤θ<π/20 \leq \theta < \pi / 20≤θ<π/2

Let $x = 2 \tan \theta$ $0 \leq \theta < \pi / 2$