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

1 Answer
Shwetank Mauria
Apr 14, 2017

#sintheta=5/13#, #costheta=-12/13#, #tantheta=-5/12#, #cottheta=-12/5#,
#sectheta=-13/12#;and #csctheta=13/5#

Explanation:

As #csctheta=13/5#, it is positive and as #costheta# is negative, #theta# lies in Quadrant II.

Now #sintheta=1/csctheta=1/(13/5)=5/13#

#costheta=-sqrt(1-(5/13)^2)=-sqrt(1-25/169)=-sqrt(144/169)=-12/13#

#tantheta=sintheta/costheta=(5/13)/(-12/13)=-5/13×13/12=-5/12#

#cottheta=1/tantheta=1/(-5/12)=-12/5#

#sectheta=1/costheta=1/(-12/13)=-13/12#

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