Question

Question #f8838

Answer 1

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`