Original question: Given costheta=-8/9cosθ=−89, thetaθ in Quadrant III, find tanthetacottheta+cscthetatanθcotθ+cscθ
First, we notice that the question is really only asking us to find sinthetasinθ since tanthetacottheta+csctheta=(sintheta/costheta)(costheta/sintheta)+1/sintheta=1+1/sinthetatanθcotθ+cscθ=(sinθcosθ)(cosθsinθ)+1sinθ=1+1sinθ, making this problem a lot less complicated than it seems.
We also know that since thetaθ is in Quadrant III, both costhetacosθ and sinthetasinθ are negative.
Consider Pythagorean's identity, sin^2theta+cos^2theta=1sin2θ+cos2θ=1.
Since we know that cos(theta)=-8/9cos(θ)=−89, we can plug in this value into Pythagorean's identity to find sin(theta)sin(θ):
sin^2theta+(-8/9)^2=1sin2θ+(−89)2=1
sin^2theta+64/81=1sin2θ+6481=1
sin^2theta=17/81sin2θ=1781
sintheta=+-sqrt(17/81)=+-sqrt(17)/9sinθ=±√1781=±√179
However, since we know that sinthetasinθ is in Quadrant III and must be negative, then sintheta=-sqrt(17)/9sinθ=−√179
Finally, since we are solving for 1+csctheta1+cscθ or 1+1/sintheta1+1sinθ, our final answer is 1+1/(-sqrt(17)/9)=1-9/sqrt(17)=(17-9sqrt(17))/171+1−√179=1−9√17=17−9√1717.
Answer: (17-9sqrt(17))/1717−9√1717