#(5pi)/6# has a reference angle of #pi/6#, so all its trig values will be related to the values for #pi/6#.
Also notice that #(5pi)/6# is located in the second quadrant, so...
#"SINE"# will be positive.
#"COSINE"# will be negative.
#sin(pi/6)=1/2#, so #color(blue)(sin((5pi)/6)=1/2#.
#cos(pi/6)=sqrt3/2#, so #color(blue)(cos((5pi)/6)=-sqrt3/2#
#"TANGENT"# is just #"SINE"# divided by #"COSINE"#:
#color(blue)(tan((5pi)/6))=(1/2)/(-sqrt3/2)=1/2(-2sqrt3)=-1/sqrt3color(blue)(=-sqrt3/3#
To figure out the other three, know that
#"SECANT"# is the reciprocal of #"COSINE"#,
#"COSECANT"# is the reciprocal of #"SINE"#, and
#"COTANGENT"# is the reciprocal of #"TANGENT"#.
#color(blue)(sec((5pi)/6))=1/(-sqrt3/2)=-2/sqrt3color(blue)(=-(2sqrt3)/3#
#color(blue)(csc((5pi)/6))=1/(1/2)color(blue)(=2#
#color(blue)(cot((5pi)/6))=1/(sqrt3/3)=3/sqrt3color(blue)(=sqrt3#
