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I thought of trying to work backwards...

$\dfrac{\sin\alpha}{\cos\alpha} + \dfrac{\cos\alpha}{\sin\alpha}\geq 2$

$\dfrac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha \cos\alpha} \geq 2$

$\dfrac{1}{\sin\alpha \cos\alpha} \geq 2$

$\sin\alpha \cos\alpha \geq \dfrac{1}{2}$

$2\sin\alpha \cos\alpha \geq 1$

$\sin 2\alpha \geq 1$

So since alpha is acute am I'm allowed to assume the final line and obtain the proof? Or am I taking the wrong and/or difficult approach?

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You will get $\sin(\alpha) \cos(\alpha) \leq \dfrac12$, when you take the reciprocal of the previous inequality (here you need to make use of the fact that $\alpha$ is acute). Rest seem fine. – user17762 Dec 17 '13 at 3:29

Here is an easier approach. For $\alpha =0 $ or $\pi/2$, the left hand side makes no sense. Hence, we can take $\alpha \in (0,\pi/2)$. We then have $\tan(\alpha)$ to be positive. Now by AM-GM, for $a,b \in \mathbb{R}^+$, we have $$a+b \geq 2\sqrt{ab}$$ Hence, we have $$\tan(\alpha) + \cot(\alpha) = \tan(\alpha) + \dfrac1{\tan(\alpha)} \geq 2 \cdot \sqrt{\tan(\alpha) \cdot \dfrac1{\tan(\alpha)}} = 2$$

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You've got it almost correct, but there's a mistake: You should conclude that $$\sin{\alpha} \cos{\alpha} \le \frac 1 2$$ not the other way around.

Here's a very different approach: The function $f(x) = \tan{x} + \cot{x}$ blows up at $0$ and $\frac{\pi}{2}$, but is differentiable in between these points. We then have

$$0 = f'(x) = \sec^2{x} - \csc^2{x}$$

Using the definitions of $\sec$ and $\csc$ together with the domain of definition, it's easy to show that this occurs at $x = \frac{\pi}{4}$, and then to establish that this is, in fact, a minimum. The minimum is then

$$f\left(\frac{\pi}{4}\right) = 1 + 1 = 2$$ as desired.

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This does not require trigonometry, if you have differential calculus and optimization.

When $x$ is positive, $x+\frac{1}{x}$ is always greater than or equal to 2 through calculus. The derivative of this expression is $1-\frac{1}{x^2}$, and the second derivative is $\frac{1}{x^3}$. So at $x=1$, there is a minimum. Specifically, $1+\frac11=2$. This is a minimum for $x+\frac1x$ over all of $(1,\infty)$, so if $\alpha$ is acute, we have $x=\tan\alpha$ in the range where this applies.

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Your idea is correct.

I am sure by $\sin 2\alpha \geq 1$ you actually mean $\sin 2\alpha \leq 1$..

and rest of things are fine....

as $\alpha $ is acute, $0\leq 2\alpha<180$ we would have $\sin 2\alpha \leq 1$ (why ??)

and you can move upwards....

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