To evaluate [tex]\(\tan\left(\sin^{-1}\left(-\frac{4}{5}\right)\right)\)[/tex], we need to recall some important trigonometric concepts and identities.
First, let's denote the angle [tex]\(\theta\)[/tex], such that
[tex]\[
\sin \theta = -\frac{4}{5}
\][/tex]
This places [tex]\(\theta\)[/tex] in either the third or fourth quadrant, where the sine function is negative. For the range of the inverse sine function, [tex]\(\theta\)[/tex] will be in the fourth quadrant, where the sine is negative and cosine is positive.
To find [tex]\(\cos \theta\)[/tex], we use the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Substituting [tex]\(\sin \theta\)[/tex]:
[tex]\[
\left(-\frac{4}{5}\right)^2 + \cos^2 \theta = 1
\][/tex]
[tex]\[
\frac{16}{25} + \cos^2 \theta = 1
\][/tex]
Solving for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\cos^2 \theta = 1 - \frac{16}{25}
\][/tex]
[tex]\[
\cos^2 \theta = \frac{25}{25} - \frac{16}{25}
\][/tex]
[tex]\[
\cos^2 \theta = \frac{9}{25}
\][/tex]
Taking the positive square root (since [tex]\(\theta\)[/tex] is in the fourth quadrant where cosine is positive):
[tex]\[
\cos \theta = \frac{3}{5}
\][/tex]
Now, we can find [tex]\(\tan \theta\)[/tex], using the definition of tangent as the ratio of sine and cosine:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
[tex]\[
\tan \theta = \frac{-\frac{4}{5}}{\frac{3}{5}}
\][/tex]
[tex]\[
\tan \theta = \frac{-4}{3}
\][/tex]
Hence, the value of [tex]\(\tan\left(\sin^{-1}\left(-\frac{4}{5}\right)\right)\)[/tex] is:
[tex]\[
-\frac{4}{3}
\][/tex]
