Original question: Given #costheta=-8/9#, #theta# in Quadrant III, find #tanthetacottheta+csctheta#
First, we notice that the question is really only asking us to find #sintheta# since #tanthetacottheta+csctheta=(sintheta/costheta)(costheta/sintheta)+1/sintheta=1+1/sintheta#, making this problem a lot less complicated than it seems.
We also know that since #theta# is in Quadrant III, both #costheta# and #sintheta# are negative.
Consider Pythagorean's identity, #sin^2theta+cos^2theta=1#.
Since we know that #cos(theta)=-8/9#, we can plug in this value into Pythagorean's identity to find #sin(theta)#:
#sin^2theta+(-8/9)^2=1#
#sin^2theta+64/81=1#
#sin^2theta=17/81#
#sintheta=+-sqrt(17/81)=+-sqrt(17)/9#
However, since we know that #sintheta# is in Quadrant III and must be negative, then #sintheta=-sqrt(17)/9#
Finally, since we are solving for #1+csctheta# or #1+1/sintheta#, our final answer is #1+1/(-sqrt(17)/9)=1-9/sqrt(17)=(17-9sqrt(17))/17#.
Answer: #(17-9sqrt(17))/17#
