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This function is undefined at 0, i.e. we have to consider it on `(-pi, 0)` and `(0, pi)` separately.
To determine concavity, we have to find the second derivative and its sign.
`f'(x) = 1 - 2(cosx)/(sin^2x),`
`f''(x) = -2*(-sinx*sin^2x - cosx*2sinx*cosx)/(sin^4x) =`
`= 2(sin^2x+2cos^2x)/(sin^3x) = 2(1+cos^2x)/(sin^3x).`
This is has the same sign as sinx, namely "-" for x in `(-pi, 0)` and "+" in` (0, pi).`
So the answer is: y is concave downwards on `(-pi, 0)` and is concave upwards on `(0, pi).`
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