Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where $0

Question

Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where $0 < x < \frac { \pi } { 2 }$ . $\sqrt { x ^ { 2 } - 16 } , x = 4 \sec \theta$ ​
A)4 tan θ
B)16 sec θ
C)- 4 tan θ
D)4 sec θ
E)16 tan θ

Quizplus · Accepted Answer

Given $x = 4\sec\theta$, substitute $x$ in $\sqrt{x^2 - 16}$ to get $\sqrt{(4\sec\theta)^2 - 16} = \sqrt{16\sec^2\theta - 16}$. Factoring out 16 gives $\sqrt{16(\sec^2\theta - 1)}$. Using the trigonometric identity $\sec^2\theta - 1 = \tan^2\theta$, we get $\sqrt{16\tan^2\theta} = 4\tan\theta$.

Use the Trigonometric Substitution to Rewrite the Algebraic Expression as a Trigonometric