Write each expression as an algebraic (nontrigonometric) expression in $u,u⩾0$.
$$\sin (2{\sec }^{-1}\frac{u}{6})$$
$$\sin (2{\sec }^{-1}\frac{u}{6})=$$
(Simplify your answer.)

Step 1 ：Given the expression $\sin (2{\sec }^{-1}\frac{u}{6})$, we need to convert it into an algebraic expression.

Step 2 ：First, we simplify the inner function, which is the inverse secant function. The secant function is the reciprocal of the cosine function, so the inverse secant function can be written in terms of the inverse cosine function. We can write ${\sec }^{-1}(x)$ as ${\cos }^{-1}(1/x)$ for $x$ in the interval $[-\mathrm{\infty },-1]\cup [1,\mathrm{\infty }]$. In our case, $x=u/6$. So, we can write ${\sec }^{-1}(u/6)$ as ${\cos }^{-1}(6/u)$.

Step 3 ：Next, we simplify the sine function. The sine of twice an angle can be written in terms of the cosine of the angle using the double-angle formula for sine, which is $\sin (2x)=2\sin (x)\cos (x)$. So, we can write $\sin (2{\cos }^{-1}(6/u))$ as $2\sin ({\cos }^{-1}(6/u))\cos ({\cos }^{-1}(6/u))$.

Step 4 ：Finally, we simplify the sine and cosine of the inverse cosine function. The sine of the inverse cosine function can be written as $\sqrt{1-x^2}$, and the cosine of the inverse cosine function is $x$. So, we can write $2\sin ({\cos }^{-1}(6/u))\cos ({\cos }^{-1}(6/u))$ as $2\sqrt{1-(6/u{)}^2}(6/u)$.

Step 5 ：This expression can be further simplified by multiplying the numerator and the denominator by $u$. This will eliminate the fraction inside the square root and the fraction outside the square root.

Step 6 ：The final algebraic expression equivalent to $\sin (2{\sec }^{-1}\frac{u}{6})$ is $\boxed{{\displaystyle 12\sqrt{1-\frac{36}{u^2}}}}$.