we have the expression
[tex]sin\left(tan^{-1}\left(-5/12\right)\right)[/tex]Let
x ----> the measure of the given angle in degrees
we have that
[tex]\begin{gathered} tan(x)=-\frac{5}{12} \\ so \\ sin(tan^{-1}(-\frac{5}{12})=sinx \\ \end{gathered}[/tex]Remember the identity
[tex]\begin{gathered} tan^2x+1=sec^2x \\ substitute\text{ given value} \\ If\text{ the value of the tangent is negative the angle x lies on the II quadrant or IV quadrant} \\ (-\frac{5}{12})^2+1=sec^2x \\ sec^2x=\frac{25}{144}+1 \\ \\ sec^2x=\frac{169}{144} \\ \\ secx=\pm\frac{13}{12} \\ \\ cosx=\pm\frac{12}{13} \end{gathered}[/tex]Find out the value of sine of the angle x
[tex]\begin{gathered} sin^2x+cos^2x=1 \\ sin^2+(\frac{12}{13})^2=1 \\ \\ sin^2x=1-\frac{144}{169} \\ sin^2x=\frac{25}{169} \\ \\ sinx=\pm\frac{5}{13} \end{gathered}[/tex]