Question #f8838

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Answer 1 · 2017-02-23T14:45:00

Since #sinx =2/3->+ve and cosx <0->-ve#

the angle x should be II quadrant

So #cosx=-sqrt(1-sin^2x)=-sqrt(1-4/9)=-sqrt5/3#

#secx=1/cosx=-3/sqrt5#

#cscx =1/sinx=3/2#

#tanx=sinx/cosx=(2/3)/(-sqrt5/3)=-2/sqrt5#

#cotx=1/tanx=-sqrt5/2#