For a ∈[π, 2 π] and n ∈ Z, the critical points of f(x)=1/3 sin a tan…

We have,

$f'(x)=\sin a+\tan ^2 x \sec ^2 x+(\sin a-1) \sec ^2 x$

$\Rightarrow f'(x)=\left(\sin a \tan ^2 x+\sin a-1\right) \sec ^2 x$

At critical points, we must have

$f'(x)=0$

$\Rightarrow \sin a \tan ^2 x+\sin a-1=0$            [∵  sec²x ≠ 0 for any x ∈ R]

$\Rightarrow \tan ^2 x=\frac{1-\sin a}{\sin a}$

Now,

$a \in[\pi, 2 \pi]$

$\Rightarrow \frac{1-\sin a}{\sin a}<0 $

$\Rightarrow \tan ^2 x=\frac{1-\sin a}{\sin a}$ does not have solution in R.

Hence, f(x) has no critical points.