the given #cot theta# at the third quadrant x is negative and y is negative.
The formulas:
#tan theta=1/cot theta#
#csc theta=+-sqrt(cot^2 theta+1)#
#sin theta=1/csc theta=1/(+-sqrt(cot^2 theta+1))#
#cos theta=sin theta*cot theta=cot theta/(+-sqrt(cot^2 theta+1))#
#sec theta=1/cos theta=(+-sqrt(cot^2 theta+1))/cot theta#
Compute the values of the other remaining trigonometric functions
Start with #cot theta=(-sqrt(11))/(-5)#
#tan theta=1/cot theta=1/((-sqrt(11))/(-5))=5/sqrt(11)=(5sqrt11)/11#
#csc theta=+-sqrt(cot^2 theta+1)=+-sqrt(((-sqrt(11))/(-5))^2+1)=+-sqrt(11/25+1)=-6/5#
#sin theta=1/csc theta=1/(+-sqrt(cot^2 theta+1))=1/(+-sqrt(((-sqrt(11))/(-5))^2+1))=1/(+-sqrt((11/25+1)))=-5/6#
#cos theta=sin theta*cot theta=cot theta/(+-sqrt(cot^2 theta+1))((-sqrt(11))/(-5))/(+-sqrt(((-sqrt(11))/(-5))^2+1))=((-sqrt(11))/(-5))/(+-6/5)=-sqrt(11)/6#
#sec theta=1/cos theta=(+-sqrt(cot^2 theta+1))/cot theta=+-sqrt(((-sqrt(11))/(-5))^2+1)/((-sqrt(11))/(-5))=(-6/5)/((-sqrt(11))/(-5))=(-6)/sqrt(11)=(-6sqrt11)/11#
God bless....I hope the explanation si useful.