for tan(cos^(-1)(x/7))

Explanation:

Step1: Let $\theta=\cos^{-1}(\frac{x}{7})$

So, $\cos\theta = \frac{x}{7}$.

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

We get $\sin\theta=\sqrt{1 - \cos^{2}\theta}=\sqrt{1 - (\frac{x}{7})^{2}}=\sqrt{1-\frac{x^{2}}{49}}=\frac{\sqrt{49 - x^{2}}}{7}$ (assuming $\theta$ is in the range where $\sin\theta\geq0$).

Step3: Recall the definition of tangent

$\tan\theta=\frac{\sin\theta}{\cos\theta}$. Substituting $\sin\theta=\frac{\sqrt{49 - x^{2}}}{7}$ and $\cos\theta=\frac{x}{7}$, we have $\tan\theta=\frac{\frac{\sqrt{49 - x^{2}}}{7}}{\frac{x}{7}}=\frac{\sqrt{49 - x^{2}}}{x}$ ($x\neq0$). Since $\theta = \cos^{-1}(\frac{x}{7})$, then $\tan(\cos^{-1}(\frac{x}{7}))=\frac{\sqrt{49 - x^{2}}}{x}$ ($x\neq0$).

Answer:

$\frac{\sqrt{49 - x^{2}}}{x}$ ($x\neq0$)