`(17pi)/12 = (9pi)/4 - (5pi)/6` Find the exact values of the sine, cosine, and tangent of the angle.

`sin(u-v)=sin(u)cos(v)-cos(u)sin(v)`

`sin((9pi)/4-(5pi)/6)=sin((9pi)/4)cos((5pi)/6)-cos((9pi)/4)sin((5pi)/6)`

`sin((9pi)/4-(5pi)/6)=(sqrt2/2)(-sqrt3/2)-(sqrt2/2)(1/2)=-sqrt2/4(sqrt3+1)` 

`cos(u-v)=cos(u)cos(v)+sin(u)sin(v)`

`cos((9pi)/4-(5pi)/6)=cos((9pi)/4)cos((5pi)/6)+sin((9pi)/4)sin((5pi)/6)`

`cos((9pi)/4-(5pi)/6)=(sqrt2/2)(-sqrt3/2)+(sqrt2/2)(1/2)=-sqrt2/4(sqrt3-1)`

`tan(u-v)=(tan(u)-tan(v))/(1+tan(u)tan(v))`

`tan((9pi)/4-(5pi)/6)=(1-(-sqrt3/3))/(1+(1)(-sqrt3/3))=((3+sqrt3)/3)/((3-sqrt3)/3)=(3+sqrt3)/(3-sqrt3)`

After rationalizing the denominator the answer is `2+sqrt3.`