Animated Solution for Mathematics - Trigonometry: For a positive integer $n$, let $f_n(\theta )=(\tan \frac{\theta }{2})(1+\sec \theta )(1+\sec 2\theta )(1+\sec 4\theta )\dots (1+\sec 2^n\theta )$. Then

We are given the function:

The secret lies in the angles. They are doubling: $\theta ,2\theta ,4\theta ,\dots ,2^n\theta$. This is the signature of a telescoping product.

Let us focus on the very first interaction: $\left(\tan \frac{\theta }{2}\right)(1+\sec \theta )$.

We know that $\sec \theta =\frac{1}{\cos \theta }$, so the bracket becomes:

Using the half-angle identity $1+\cos \theta =2{\cos }^2\frac{\theta }{2}$, we rewrite the bracket as:

Now, multiply this by $\tan \frac{\theta }{2}=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}$. The $\cos \frac{\theta }{2}$ in the denominator cancels with one of the $\cos \frac{\theta }{2}$ terms in the numerator, leaving us with:

That numerator is the classic double-angle identity for sine. It simplifies to $\sin \theta$. Thus, the entire expression collapses to:

Now that we have $\tan \theta$, we multiply it by the next bracket, $(1+\sec 2\theta )$.

Following the exact same logic:

Multiplying this by our current result, $\tan \theta =\frac{\sin \theta }{\cos \theta }$, we get:

Do you see the pattern? Every time we multiply by a new bracket, the angle of the tangent function doubles. By induction, after $n$ such multiplications, the entire product simplifies to:

Now, we simply test the options. For Option A, $f_2(\pi /16)$, we have $n=2$ and $\theta =\pi /16$.

Our formula gives:

For Option B, $f_3(\pi /32)$, we have $n=3$ and $\theta =\pi /32$.

Our formula gives:

The pattern holds for all options! Each one reduces to $\tan (\pi /4)$, which is 1. This problem is a beautiful reminder that in mathematics, no matter how complex the expression, there is always an underlying structure waiting to be discovered.