For $0 < x < π/2$, show that $\cot(x) + \tan(x) ≥ 2$

Question

So, I know this sounds simple. I just can't seem to find a way to prove it without using drawings.

I used trig identities to find a different way to express $$\cot(x) + \tan(x) =\frac{ 1}{\sin(x)\cos(x)}$$ Clearly there are no roots.

Could anyone give me a hint on which direction I should lead?

Answer 1

By AM-GM $$\tan{x}+\cot{x}\geq2\sqrt{\tan{x}\cot{x}}=2.$$

Also, $$\tan{x}+\cot{x}=\frac{\sin^2x+\cos^2x}{\sin{x}\cos{x}}=\frac{2}{\sin2x}\geq2$$ because $0<\sin2x\leq1$ for $0<x<\frac{\pi}{2}$.

Answer 2

Let $\tan(x)=y$. Then we have $\cot(x)=1/y$. Then your sum is $y+1/y$, which is at least 2 when $y\geq 0$. Finally see that $tan(x)\geq 0$ on the desired interval.

Answer 3

set $$\tan(x)=t$$ then we have $$\cot(x)=\frac{1}{t}$$ and we Need to prove that $$\frac{1}{t}+t\geq 2$$ since $t>0$ in the given interval we can multiply by $t>0$ and we get $$t^2+1\geq 2t$$ or $$(t-1)^2\geq 0$$ which is true.

Answer 4

You actually are really close already. \begin{align} \cot x + \tan x &=\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}\\ &=\frac{1}{\sin x \cos x}\\ &=\frac{2}{\sin 2x}\\ \end{align} Since $0<\sin 2x < 1$ for $0<x<1$, $$\frac{2}{\sin 2x}>2$$ Hence, $$\cot x + \tan x>2$$ for $0<x<1$.

Answer 5

Note that for all $x\in[0,\pi]$ we have, $$0\le \sin x\le 1 \implies \frac{1}{\sin x}\ge 1$$ Hence,

$$\tan{x}+\cot{x}=\frac{\sin^2x+\cos^2x}{\sin{x}\cos{x}}=\frac{2}{2\sin{x}\cos{x}} =\frac{2}{\sin2x}\ge2$$