Question 607039
You must have mixed part of a question with something else.
Or else this is a trick question.
{{{pi/2}}} is {{{180^o}}} in degrees, and
{{{sin(pi/2)=0}}} , {{{cos(pi/2)=-1}}} , and {{{tan(pi/2)=0}}} ,
regardless of the value {{{x}}}, or {{{csc(x)}}} .
 
On the other hand, {{{csc(x)=1/sin(x)}}} , so
if {{{csc(x)=1/sin(x)=6}}} , then {{{sin(x)=1/6}}}
and since {{{(sin(x))^2+(cos(x))^2=1}}}, then
{{{(cos(x))^2=1-(1/6)^2=1-1/36=35/36}}}
And if we also know that {{{90^o<x<180^o}}}, we know that {{{cos(x)<0}}}
so that would mean {{{cos(x)=-sqrt(35/36)=-sqrt(35)/sqrt(36)=-sqrt(35)/6}}}
Then as {{{tan(x)=sin(x)/cos(x)}}} , we would have
{{{tan(x)=(1/6)/(-sqrt(35)/6)=-(1/6)*(6/sqrt(35))=-1/sqrt(35)=-sqrt(35)/35}}}