If csc θ =-4/3 and cos θ >0, in which quadrant does θ terminate?

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Answer 1 · 2018-03-30T09:01:50

# theta in [(3pi)/2,2pi]=>4^(th)# quadrant

Here,

#csctheta=-4/3=>sintheta=-3/4 <0#

Hence,

#sintheta<0 and costheta>0#

#=> theta in [(3pi)/2,2pi]=>4^(th)# quadrant

Answer 2 · 2018-03-30T09:07:07

IV

Identity.

#color(red)bb(cscx=1/sinx)#

#cscx=1/sinx=-4/3#

#:.#

#sinx=-3/4#

#x=arcsin(sinx)=arcsin(-3/4)~~-0.84806#

This is in the IV quadrant.

#cos(arcsin(-3/4))>0#

#:.#

#cos(-x)=cos(x)#