Question

find the exact value of the expression given that secx = (3/2), cscy = 3, and x and y are in Quadrant I

cos(x - y)

Answer 1

\(\bf {\color{blue}{ cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies \cfrac{a}{c}\qquad
sin(\theta)=\cfrac{opposite}{hypotenuse}\implies \cfrac{b}{c}}}\\ \quad \\ \quad \\

sec(x)=\cfrac{3}{2}\qquad csc(y)=3=\cfrac{3}{1}\\ \quad \\ \quad \\
sec(x)=\cfrac{3}{2}\implies \cfrac{1}{cos(x)}=\cfrac{3}{2}\implies {\color{blue}{ \cfrac{2}{3}=cos(x)}}\implies \cfrac{a=2}{c=3}\\ \quad \\
csc(y)=\cfrac{3}{1}\implies \cfrac{1}{sin(y)}=\cfrac{3}{1}\implies {\color{blue}{ \cfrac{1}{3}=sin(y)}}\implies \cfrac{b=1}{c=3}\)

Answer 2

now, for the angle "x", you need the "b" component, that is the opposite side
for that you can use the pythagorean theorem \(\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}={\color{blue}{ b}}\)

for the angle "y" you need the "a" component, that is the adjacent side
for that you use the pythagorean theorem as well \(\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}={\color{blue}{ a}}\)

notice that the pythagorean theorem uses a radical, which can give a positive or negative value, both valid

so which one do we pick?
well, we know that "x" and "y" are in Quadrant I, that makes a,b and c, all positive

and to expand cos(x-y), you'd need both cosines and sines