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