find the exact value of tan(sin^(-1)(12/13)).

Explanation:

Step1: Let $\theta=\sin^{-1}(\frac{12}{13})$.

This means $\sin\theta=\frac{12}{13}$, and $\theta$ is in the range $[-\frac{\pi}{2},\frac{\pi}{2}]$.

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

We get $\cos\theta=\sqrt{1 - \sin^{2}\theta}=\sqrt{1-(\frac{12}{13})^{2}}=\sqrt{\frac{169 - 144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$ (since $\cos\theta\geq0$ for $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$).

Step3: Recall the definition of tangent.

$\tan\theta=\frac{\sin\theta}{\cos\theta}$, substituting $\sin\theta=\frac{12}{13}$ and $\cos\theta=\frac{5}{13}$, we have $\tan\theta=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$.

Answer:

$\frac{12}{5}$