How do you find the exact value of the five remaining trigonometric function given #sectheta=-sqrt10# and the angle is in standard position in quadrant III?

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Answer 1 · 2016-12-26T08:34:39

#costheta=-1/sqrt(10)#;
#sintheta=-3/sqrt(10)#;
#tantheta=3#

#cottheta=1/3#

#csctheta=-sqrt(10)/3#

Since

#costheta=1/sectheta#

then

#costheta=1/-sqrt(10)=-1/sqrt(10)#

Since the angle is in standard position in quadrant III, and

#sintheta=+-sqrt(1-cos^2theta)#

then #sintheta=color(red)-sqrt(1-(1/(-sqrt(10)))^2)=-sqrt(1-1/10)=-sqrt(9/10)=-3/sqrt(10)#

So

#tantheta=sintheta/costheta=(-3/cancel(sqrt(10)))/-(1/cancelsqrt(10))=3#

#cottheta=1/tantheta=1/3#

#csctheta=1/sintheta=1/-sqrt(9/10)=-sqrt(10)/3#