1 Answer

Shiznatix · Answer 1 · 2023-08-16T05:41:01+0000

Since angle Cita lies on the 2nd quadrant, then

The value of tan Cita is negative

Let us use the identites

$\begin{gathered} sin^2\theta+cos^2\theta=1\rightarrow(1) \\ tan\theta=(sin\theta)/(cos\theta)\rightarrow(2) \end{gathered}$

Since sin Cita = 4/5, then substitute it in (1)

$\begin{gathered} ((4)/(5))^2+cos^2\theta=1 \\ (16)/(25)+cos^2\theta=1 \end{gathered}$

Subtract 16/25 from each side

$\begin{gathered} (16)/(25)-(16)/(25)+cos^2\theta=1-(16)/(25) \\ cos^2\theta=(9)/(25) \end{gathered}$

Take a square root for both sides

$\begin{gathered} √(cos^2\theta)=\pm\sqrt{(9)/(25)} \\ cos\theta=\pm(3)/(5) \end{gathered}$

Since Cita lies in the 2nd quadrant, then its cosine is negative

$cos\theta=-(3)/(5)$

Now, use identity (2) to find tan Cita

$\begin{gathered} tan\theta=((4)/(5))/(-(3)/(5)) \\ tan\theta=(4)/(5)*-(5)/(3) \\ tan\theta=-(4)/(3) \end{gathered}$

The answer is

tan Cita = -4/3

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