Question #978dc

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Answer 1 · 2016-08-24T06:05:52

Given that#(-sqrt3,1)# is the cartesian coordinate of the point on the terminal side of the angle #theta#.It is in the third quadrant.

Here #x=-sqrt3 and y =1 #

So lrngth of the terminal side
#r=sqrt(x^2+y^2)=sqrt((-sqrt3)+1^2)=2#

#"Here in respect of "theta#
# x->"adjacent"#

#y->"opposite"#

#r->"hypotenuse"#

#sintheta=y/r=1/2#

#costheta=x/r=-sqrt3/2#

#tantheta=y/x=1/(-sqrt3)-1/sqrt3#

#csctheta=r/y=2/1=2#

#sectheta=r/x=-2/sqrt3#

#cottheta=x/y=-sqrt3/1#