How do you use #tantheta=4# to find #costheta#?

You could use the identity #sec^2 theta = tan^2theta +1# to find that in this case #sec^2 theta = 16+1=17#, so that #sec theta = pm sqrt{17}# Since #sec theta = 1/cos theta#, this tells us that

#cos theta = pm 1/sqrt{17}#

Unfortunately the sign of #cos theta# can not be found from the value of #tan theta#. Since #tan theta # is positive in this case, the angle #theta# may lie in either the first or the third quadrant. If the angle is in the first quadrant, then #cos theta = 1/sqrt{17}#. If, on the other hand, it is in the third quadrant, then #cos theta = -1/sqrt{17}#

If #tan(theta)=4#
then the ratio of the opposite side to the adjacent side is #4:1#

Two possibilities exist depending upon whether #theta# is in Quadrant I or Quandrant III; as indicated below:

In either case the relative length of the hypotenuse is given by the Pythagorean Theorem as
#color(white)("XXX")h=sqrt(4^2+1^2)=sqrt(17)#

Since
#color(white)("XXX")cos(theta)="adjacent"/("hypotenuse")#
for this case
#color(white)("XXX")cos(theta)=1/sqrt(17)color(white)("xx")"or"color(white)("xx")(-1)/sqrt(17)#