The question is as such
If the sum of the first $n$ terms of an arithmetic progression is denoted by $S_n$ and
$$S_n = 6n\sec^2 \theta + n(n-1)(\sin^2 \theta(4 + \tan^2 \theta) + \cos^2 \theta(4 + \cot^2 \theta))$$
where $\theta \in (0, \frac{\pi}{2})$. Then the minimum value of the common difference of the arithmetic progression is
(a) 10, (b) 12, (c) 14, or (d) 16,
could not find a similar question on google, I tried to break up the $4 + \tan^2 \theta$ and $4 + \cot^2 \theta$ in the brackets into $\sec^2 \theta + 3$ and $\csc^2 \theta + 3$ and then get somewhere however could not
Any help would be appreciated
Hint For an arithmetic progression $a_i = k i + a_0$ (our aim is to minimize $k$), the sum of the first $n$ terms is $$\sum_{i = 0}^{n - 1} (k i + a_0) = n (n - 1) \cdot \frac12 k + n \cdot a_0 .$$ So, $$\frac12 k = \sin^2 \theta (4 + \tan^2 \theta) + \cos^2 \theta (4 + \cot^2 \theta). $$ In terms of $u := \sin^2 \theta$, which takes on precisely the values $(0, 1)$, $$\frac12 k = 1 + \frac{1}{u (1 - u)}.$$