1) If sin alpha = -4/5, and cos alpha>0, find tan alpha and sec alpha.2) If cot x = -3/2 and sec x<0, find sin x and cos x.3) If cos theta = 0.54, find sin(theta-pi/2).4) If cot x = -0.18, find tan(x-pi/2).Please explain/show steps, I'm very lost on how to do these!

1)

cos(α) > 0, is just another way to say cos(α) is positive, and therefore, the adjacent side is positive then.

now, on -4/5, keep in mind that the hypotenuse is just a radius unit, and therefore is never negative, so in -4/5, the negative has to be the -4.

[tex]\bf sin(\alpha)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm 3=a\implies \stackrel{cos(\alpha)\ \textgreater \ 0}{+3=a}\\\\ -------------------------------[/tex]

[tex]\bf tan(\alpha)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{adjacent}{3}}\qquad \qquad sec(\alpha)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{3}}[/tex]



2)

[tex]\bf cot(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{opposite}{2}}\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{(-3)^2+2^2}\implies c=\sqrt{13}\\\\ -------------------------------[/tex]

[tex]\bf sin(x)=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{\sqrt{13}}}\quad \stackrel{rationalizing~it}{\implies }\quad \cfrac{2}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{2\sqrt{13}}{13} \\\\\\ cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{\sqrt{13}}}\quad \stackrel{rationalizing~it}{\implies }\quad \cfrac{-3}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies -\cfrac{3\sqrt{13}}{13}[/tex]

hmm the sec(x) < 0, is just another way to say sec(x) is negative, and since the cosine is the reciprocal of the secant, then the cosine is also negative.  Now in the fraction -3/2, which is the negative?  since it could be -3/2 or 3/-2, well, anyhow, is the -3, because is the adjacent side, which is used by the cosine, that we know is negative.