Instead of using formulas, it'd be easier to solve it geometrically, with a right triangle.

Since #csc theta = 1/sintheta = "hypotenuse"/"opposite"=c/a = 4/3#, this means that #a# and #c# are multiples of #3# and #4#, respectively.

In other words, we have #c=4k# and #a=3k#, for a real number #k#.

By the Pythagorean theorem, #b = sqrt(c^2-a^2) = sqrt(16k^2-9k^2) = sqrt(7)*k#.

Finally, for trigonometric functions :

#sin theta = "opposite"/"hypotenuse" = a/c = 3/4#

#cos theta = "adjacent"/"hypotenuse" = b/c = sqrt7/4#

#tan theta = "opposite"/"adjacent" = a/b = 3/sqrt7#

#cot theta = 1/tan theta = b/a = sqrt7/3#

#sec theta = "hypotenuse"/"adjacent" = c/b = 4/sqrt7#.