# Minimum value of : $\left|\sin{x}+ \cos{x} + \sec{x} + \tan{x} + \cot{x} + \csc{x}\right|$ [duplicate]

Find the minimum value of :$$\left|\sin{x}+ \cos{x} + \sec{x} + \tan{x} + \cot{x} + \csc{x}\right|.$$

What I have tried is trying to simplify the expression into a $\sin{2x}$ form but it did not go well. I guess there is much simpler way to solve the problem.

I also thought of AM-GM inequality but for inequality to hold all the terms must be positive real numbers which in this case is not.

## marked as duplicate by Ng Chung Tak, Leucippus, hardmath, C. Falcon, JunivenApr 15 '17 at 2:13

• You could just rewrite everything in terms of cos(x), likely leading to a fourth-degree polynomial to maximize. – TMM Apr 14 '17 at 15:38

$$\bigg|\sin x+\cos x+\tan x+\cot x +\csc x+\sec x\bigg|$$

$$= \bigg|\sin x+\cos x+\frac{1}{\sin x\cos x}+\frac{\sin x+\cos x}{\sin x\cos x}\bigg|$$

$$= \bigg|(\sin x+\cos x)+\frac{2}{\sin 2x}+\frac{2(\sin x+\cos x)}{\sin 2x}\bigg|$$

put $\sin x+\cos = t,$ and $\sin 2x = t^2-1$ where $|t|<=\sqrt{2}$

so $$\bigg|t+\frac{2}{t^2-1}+\frac{2t}{t^2-1}\bigg| = \bigg|t-1+\frac{2}{t-1}+1\bigg|$$

let $$y = \bigg(t-1+\frac{2}{t-1}+1\bigg)$$

$\star$ If $1<t <=\sqrt{2}.$ Then using A.M G.M , we have $$\displaystyle y=t-1+\frac{2}{t-1}+1\geq 2\sqrt{2}+1$$

$\star$ If $-\sqrt{2} <=t<-1$ or $-1<t <1$ . Then using A.M G.M , $$\displaystyle -y = (1-t)+\frac{2}{1-t}>= 2\sqrt{2}-1$$

So $$y \geq 2\sqrt{2}-1$$ when $t=1-\sqrt{2}$

• How did you get $\bigg|t+\frac{2}{t^2-1}+\frac{2t}{t^2-1}\bigg| = \bigg|t-1+\frac{2}{t-1}+1\bigg|$ ? – Shreyas S Apr 14 '17 at 17:07
• partial fraction.. – Abhash Jha Apr 15 '17 at 4:48