Respuesta :

lukyo
[tex]\theta=\sin ^{-1}\left(-\dfrac{4}{5} \right )[/tex] is in the fourth quadrant. Hence,

[tex]\tan \theta<0[/tex]


From a right triangle (check out the attachment) we get

[tex]\bullet\;\;\cos \theta=\dfrac{3}{5}\\\\\\ \bullet\;\;\tan \theta=-\dfrac{4}{3}\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \tan\left(\sin^{-1} \left(-\dfrac{4}{5} \right )\right)=-\dfrac{4}{3} \end{array}}[/tex]

Ver imagen lukyo
adioabiola

Here, we are required to evaluate tan(sin^-1 (-4/5)).

tan(sin^-1 (-4/5)) = (-4/3)

Sin^-1 (-4/5) = θ

Therefore; ultimately Sin θ = (-4/5)

Since; Sine = opposite / hypothenus

Therefore, adj = √(5²) - (4²)

adj = 3.

However, since tan(sin^-1 (-4/5)) = tan θ

Therefore, tan θ = 4/3

However, since sin θ and tan θ are only negative in the fourth quarter;

Therefore, tan θ = (-4/3)

Read more:

https://brainly.com/question/21546908

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