Answer :
Using the trigonometric identities
[tex]\sec ^2(t)=\tan ^2(t)+1[/tex]We want to find out the tan(t), then we can manipulate that formula and find tan(t) in function of sec(t).
[tex]\tan ^2(t)=\sec ^2(t)-1[/tex]Now we can do square roots on both sides
[tex]\tan (t)=\pm\sqrt[]{\sec ^2(t)-1}[/tex]We know that sec(t) = 3/2, then let's put it in our formula and simplify
[tex]\begin{gathered} \tan (t)=\pm\sqrt[]{\frac{3^2}{2^2}-1} \\ \\ \tan (t)=\pm\sqrt[]{\frac{9}{4}-1} \\ \\ \tan (t)=\pm\sqrt[]{\frac{9}{4}-\frac{4}{4}} \\ \\ \tan (t)=\pm\sqrt[]{\frac{5}{4}} \end{gathered}[/tex]We can simplify and remove 4 from the square root, and we have
[tex]\tan (t)=\pm\frac{\sqrt[]{5}}{2}[/tex]But which value is correct? the positive or the negative? Now we must use the information that the problem tells us, it says that t is in the quadrant IV, the tangent in quadrant IV is negative, then
[tex]\tan (t)=-\frac{\sqrt[]{5}}{2}[/tex]