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The minimum value of $tan(x) + cot(x)$ is $-\infty$. However, if you attempt to find it this way: $$(tan(x) + cot(x))^2 = tan^2x+cot^2x+2tan(x)cot(x)$$ $$(tan(x) + cot(x))^2 = tan^2x + cot^2x+2$$ By the A.M-G.M. inequality, $$tan^2x+cot^2x \ge 2$$ So $(tan(x) + cot(x))^2$ is at least equal to $2+2=4$. So if we want the minimum value of $tan(x) + cot(x)$ we take the square root, $-2$. What's wrong with this argument?

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I realized it while typing out the question. Since I was taking the negative value of the square root, the larger the value of $(tan(x) + cot(x))^2$ is, the more negative the value I can get.

So my way of taking the minimum value of $tan^2x + cot^2x$ as $2$ is not preferred, rather I should take its maximum value, which gives both maximum and minimum values of $(tan(x) + cot(x))^2$ as $+\infty$ and $-\infty$ respectively.

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