$\cot x - \tan x = 2\cot 2x$, but the period on the left is $\pi$, while on the right it's $\pi/2$. Shouldn't the periods match?

Question

It's not difficult to show $$\cot x - \tan x = 2\cot 2x$$ The period of LHS is $T_1 = \pi$ but the period of RHS is $T_2 = \pi /2$ . Why are $T_1$ and $T_2$ not equal? So we can't use that identity for solving problems?

Answer 1

Say $f(x) = 2\cot (2x)$ and let $T_2$ be a period for $f$, then $$f(x+T_2)= f(x)\implies \cot (2x+2T_2)= \cot 2x$$

Mark $\alpha = 2x$ (so clearly $\alpha$ can be any angle) then we have $$\cot (\alpha +2T_2) = \cot (\alpha)$$

and thus $2T_2$ is a period for $\cot$ so $2T_2 = \pi$

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But $\pi/2$ is also period for $g(x)=\cot x-\tan x$: $$ g(x+\pi/2) = \cot(x+\pi/2)-\tan (x+\pi/2)$$ $$ =\cot(x-\pi/2)-\tan (x-\pi/2)$$ $$ =-\cot(\pi/2-x)+\tan (\pi/2-x)$$ $$ =-\tan(x)+\cot (x)=g(x)$$