find the exact value of $\\sin\\left(\\tan^{-1}\\left(-\\frac{5}{12}\\right)\\right)$

Explanation:

Step1: Let $\theta=\tan^{-1}(-\frac{5}{12})$

By the definition of the inverse - tangent function, $\tan\theta =-\frac{5}{12}$, and $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$. Since $\tan\theta<0$, $\theta\in(-\frac{\pi}{2},0)$.

Step2: Use the identity $\tan\theta=\frac{y}{x}$ and $r = \sqrt{x^{2}+y^{2}}$

If $\tan\theta=\frac{y}{x}=-\frac{5}{12}$, we can assume $y=- 5$ and $x = 12$. Then, by the Pythagorean theorem $r=\sqrt{x^{2}+y^{2}}=\sqrt{12^{2}+(-5)^{2}}=\sqrt{144 + 25}=\sqrt{169}=13$.

Step3: Use the definition of the sine function $\sin\theta=\frac{y}{r}$

Since $\sin\theta=\frac{y}{r}$ and $y=-5$, $r = 13$, we have $\sin\theta=\sin(\tan^{-1}(-\frac{5}{12}))=-\frac{5}{13}$.

Answer:

$-\frac{5}{13}$