Question

Question #d0dcc

Answer 1

Given `" "tanA=1/3->+ve=>cotA=3`

and `" "cosA->-ve`

So angle `Ain "III Quadrant"` where `" sinA->-ve`

So `sinA=1/csca=-1/sqrt(1+cot^2A)=-1/sqrt(1+3^2)`

`=-1/sqrt10=-sqrt10/10`