1) given cotθ = 1/2√7, find sinθ and cosθ in quadrant I

Question

2) given tanθ = √5, find secθ and cotθ in quadrant III

Answer 1

recall that
sinθ = y/r
cosθ = x/r
tanθ = y/x
r^2 = x^2+y^2

So, in QIII, where x and y are both negative,
tanθ = √5 = √5/1, so
y = -√5
x = -1
r = √6
cotθ = x/y = 1/√5
secθ = r/x = -√6

For #1, do you mean (1/2) √7 or 1/(2√7)?
In any case, follow the steps above.

Answer 2

for #1 i mean (1/2) √7

which 1/2 be the x and √7 be the y?

Answer 3

would*

Answer 4

so, cotθ = (√7)/2, meaning
x = √7
y = 2
since cotθ = x/y