Answer
$\cos(\tan^{-1}\big(\frac{x}{\sqrt{9-x^2}}\big)) = \frac{\sqrt{9-x^2}}{3}$
Work Step by Step
We are trying to find the value of the expression $\cos(\tan^{-1}\big(\frac{x}{\sqrt{9-x^2}}\big))$. Seeing the $\sqrt{9-x^2}$, it should resemble a side of a right triangle with a hypotenuse of $3$ and a leg of length $x$. Let $\theta$ be the angle between the side with length $\sqrt{9-x^2}$ and $3$. From here, we will try to obtain the inverse tangent term inside of the cosine. We see that $\tan(\theta) = \frac{x}{\sqrt{9-x^2}}$ and solving for $\theta$, we get $\theta = \tan^{-1}\big(\frac{x}{\sqrt{9-x^2}}\big)$. Substituting this into our original expression, we obtain $\cos(\tan^{-1}\big(\frac{x}{\sqrt{9-x^2}}\big)) = \cos(\theta) = \frac{\sqrt{9-x^2}}{3}$.
