Since angle Cita lies on the 2nd quadrant, then
The value of tan Cita is negative
Let us use the identites
[tex]\begin{gathered} sin^2\theta+cos^2\theta=1\rightarrow(1) \\ tan\theta=\frac{sin\theta}{cos\theta}\rightarrow(2) \end{gathered}[/tex]Since sin Cita = 4/5, then substitute it in (1)
[tex]\begin{gathered} (\frac{4}{5})^2+cos^2\theta=1 \\ \frac{16}{25}+cos^2\theta=1 \end{gathered}[/tex]Subtract 16/25 from each side
[tex]\begin{gathered} \frac{16}{25}-\frac{16}{25}+cos^2\theta=1-\frac{16}{25} \\ cos^2\theta=\frac{9}{25} \end{gathered}[/tex]Take a square root for both sides
[tex]\begin{gathered} \sqrt{cos^2\theta}=\pm\sqrt{\frac{9}{25}} \\ cos\theta=\pm\frac{3}{5} \end{gathered}[/tex]Since Cita lies in the 2nd quadrant, then its cosine is negative
[tex]cos\theta=-\frac{3}{5}[/tex]Now, use identity (2) to find tan Cita
[tex]\begin{gathered} tan\theta=\frac{\frac{4}{5}}{-\frac{3}{5}} \\ tan\theta=\frac{4}{5}\times-\frac{5}{3} \\ tan\theta=-\frac{4}{3} \end{gathered}[/tex]The answer is
tan Cita = -4/3
