question 4 using the trigonometric substitution, x = 8 sec θ, given 0 ≤ θ < π/2. simplify the expression…

Explanation:

Step1: Substitute $x = 8\sec\theta$

Substitute $x$ into the expression $\frac{\sqrt{x^{2}-64}}{x}$, we get $\frac{\sqrt{(8\sec\theta)^{2}-64}}{8\sec\theta}$.

Step2: Simplify the numerator

First, expand $(8\sec\theta)^{2}-64$: $(8\sec\theta)^{2}-64 = 64\sec^{2}\theta - 64=64(\sec^{2}\theta - 1)$. Since $\sec^{2}\theta-1=\tan^{2}\theta$, then $\sqrt{64(\sec^{2}\theta - 1)}=\sqrt{64\tan^{2}\theta}$. Given $0\leq\theta<\frac{\pi}{2}$, $\tan\theta\geq0$, so $\sqrt{64\tan^{2}\theta}=8\tan\theta$.

Step3: Simplify the whole - expression

The original expression becomes $\frac{8\tan\theta}{8\sec\theta}$. Since $\sec\theta=\frac{1}{\cos\theta}$, the expression is $\frac{8\tan\theta}{8\frac{1}{\cos\theta}}$. And $\tan\theta=\frac{\sin\theta}{\cos\theta}$, so $\frac{8\frac{\sin\theta}{\cos\theta}}{8\frac{1}{\cos\theta}}=\sin\theta$.

Answer:

$\sin\theta$