Trigonometry Solution – Problem 15: Find the minimum positive period of

Question

Period of Tangent Composition

Find the minimum positive period of $f(x) = \frac{2\tan x}{1-\tan^2 x}$ .

Recognize the double-angle tangent formula:

$f(x) = \frac{2\tan x}{1-\tan^2 x} = \tan 2x$

where $x \neq \frac{k\pi}{2}$ and $\tan x \neq \pm 1$ (i.e., $x \neq \frac{\pi}{4} + \frac{k\pi}{2}$ ).

The period of $\tan 2x$ is:

$T = \frac{\pi}{2} \cdot 2 = \pi$

$\pi$

Translate the question into conditions (2 pts): domain/range/period/symmetry/monotonicity constraints are written correctly.
Key transformation or rewrite (2 pts): rewrite the function into a standard form (e.g. amplitude/phase/period) or reduce using identities.
Correct interval/parameter reasoning (2 pts): derive the correct inequalities or argument interval.
Final answer (1 pt): state the final set / interval / period clearly.

Listing properties ("periodic", "even") without applying them to the given function.