f′(x)=sina⋅tan2xsec2x+(sina−1)sec2x =(sina⋅tan2x+sina−1)sec2x
At critical points, f′(x)=0 ⇒sinatan2x+sina−1=0 [∵sec2x=0 for any x∈R] ⇒tan2x=sina1−sina
Since a ∈[π,2π] ∴sina1−sina<0 ∴ the equation tan2x=sina1−sina does not have a solution in R. Hence f(x) has no critical points.