Question 1155696
If {{{sin(theta)=3/7}}}, And Theta¸ Is In Quadrant II, 


use {{{sin^2(theta) + cos^2 (theta)=1 }}}=> 


{{{cos^2 (theta) = 1-sin^2(theta) }}}


{{{sin^2(theta) =(3/7)^2=9/49}}}


{{{cos^2 (theta)=1- 9/49 }}}


{{{cos^2 (theta)=40/49}}}


{{{cos(theta)=sqrt(40/49)}}}


{{{cos(theta)=(2sqrt(10))/7}}}=> is negative because the angle is in the second quadrant 


{{{cos(theta)=-(2sqrt(10))/7}}}




{{{tan(theta)= sin (theta)/cos (theta)}}}


{{{tan(theta)=(3/7)/(-(2*sqrt(10))/7)}}}


{{{tan(theta)=3/(-2sqrt(10))}}}


{{{tan(theta)=3sqrt(10)/(-2sqrt(10)sqrt(10))}}}


{{{tan(theta)=3sqrt(10)/(-2*10)}}}


{{{tan(theta)=-3sqrt(10)/20}}}




{{{cot(theta)=cos(theta)/sin(theta)}}}


{{{cot (theta)=(-(2sqrt(10))/7)/(3/7)}}}


{{{cot(theta)=-(2*sqrt(10))/3}}}




{{{sec(theta)=1/cos(theta)}}}


{{{sec(theta)=1/(-2sqrt(10)/7)}}}


{{{sec(theta)=7/(-2sqrt(10))}}}


{{{sec(theta)=-7sqrt(10)/20}}}