Question

(tan(sin-1(-5/13)) : FIND THE EXACT VALUE:)

Answer 1

\(\bf tan\left(sin^{-1}\left(-\frac{5}{13}\right)\right)\\ \quad \\ \quad \\
\color{green}{sin^{-1}\left(-\frac{5}{13}\right)} = \theta\quad thus \quad sin(\theta) = -\cfrac{5}{13}\implies \cfrac{b}{c}\\ \quad \\
\color{blue}{c^2= a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\implies a = \pm\sqrt{13^2-5^2}=12}\\ \quad \\
tan(\theta)= \cfrac{b}{a}\implies\cfrac{-5}{12}\qquad thus \qquad tan^{-1}\left(\frac{-5}{12}\right)\Leftrightarrow\theta \Leftrightarrow \color{green}{sin^{-1}\left(-\frac{5}{13}\right)}\\ \quad \\
tan\left(sin^{-1}\left(-\frac{5}{13}\right)\right) \Leftrightarrow tan\left(tan^{-1}\left(\frac{-5}{12}\right)\right) \)


can you see now what the tangent is?

Answer 2

now, why is "a" not -12? well, because the inverse sine function is restricted to \(\bf \frac{\pi}{2},-\frac{\pi}{2}\) and we have a negative sine, thus that means, the angle given will end up in the 4th Quadrant