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

Question

Let $x = 2 \tan \theta$ , $0 \leq \theta < \frac { \pi } { 2 }$) 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 $\sqrt{4+4\tan^2\theta} = \sqrt{4(1+\tan^2\theta)}$, which simplifies to $2\sqrt{1+\tan^2\theta}$. Since $1+\tan^2\theta = \sec^2\theta$, the expression becomes $\frac{1}{4\tan^2\theta \cdot 2\sec\theta}$. Further simplification using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sec\theta = \frac{1}{\cos\theta}$ leads to $\frac{\cos^3\theta}{8\sin^2\theta}$.

Let x=2tan⁡θx = 2 \tan \thetax=2tanθ 0≤θ<π20 \leq \theta < \frac { \pi } { 2 }0≤θ<2π​

Let $x = 2 \tan \theta$ $0 \leq \theta < \frac { \pi } { 2 }$