How do you find the remaining trigonometric functions of #theta# given #sectheta-=13/5# and #sintheta<0#?

1 Answer
Aug 18, 2017

Answer:

#"see explanation"#

Explanation:

#"since "sintheta<0" this indicates that "theta" is in the"#
#"the third or fourth quadrants"#

#•color(white)(x)costheta=1/sectheta=1/(13/5)=5/13#

#•color(white)(x)sintheta=+-sqrt(1-cos^2theta)#

#rArrsintheta=-sqrt(1-(5/13)^2)#

#color(white)(rArrsintheta)=-sqrt(144/169)=-12/13#

#•color(white)(x)csctheta=1/sintheta=1/(-12/13)=-13/12#

#•color(white)(x)tantheta=sintheta/costheta=(-12/13)/(5/13)=-12/5#

#•color(white)(x)cottheta=1/tantheta=1/(-12/5)=-5/12#

#theta" is in the fourth quadrant"#