all trigo functions

\hspace{-16}$Let $\bf{f(x)=\left|\sin x+\cos x+\tan x+\cot x+\csc x+\sec x\right|}$\\\\\\ $\bf{=\left|\sin x+\cos x+\frac{\sin^2 x+\cos^2 x}{\sin x.\cos x}+\frac{\sin x+\cos x}{\sin x.\cos x}\right|}$\\\\\\ $\bf{=\left|\left(\sin +\cos x\right)+\frac{2}{\sin 2x}+\frac{2\left(\sin x+\cos x\right)}{\sin 2x}\right|}$\\\\\\ Let $\bf{\sin x+\cos x=t\Leftrightarrow 1+\sin 2x =t^2\Leftrightarrow \sin 2x =t^2-1}$\\\\\\ and $\bf{-\sqrt{2}\leq \left(\sin x+\cos x\right)\leq \sqrt{2}\Leftrightarrow -\sqrt{2}\leq t \leq \sqrt{2}}$\\\\\\ So $\bf{f(t)=\left|t+\frac{2}{t^2-1}+\frac{2t}{t^2-1}\right|=\left|t+\frac{2}{t-1}\right|}$\\\\\\ Now Let $\bf{g(t)=t+\frac{2}{t-1}}$\\\\\\ Diff. both side w.r.to $\bf{x}$\\\\\\ $\bf{g\hspace{-10}\quad'(x)=1-\frac{2}{\left(t-1\right)^2}}$\\\\\\ Now for Max. or Min. $\bf{g\hspace{-10}\quad'(x)=0\Leftrightarrow x=1\pm \sqrt{2}}$\\\\\\ So $\bf{g(t)}$ has relative exterma at $\bf{x=\sqrt{2}\;\;,-\sqrt{2}}$ and $\bf{x=1-\sqrt{2}}$\\\\\\

\hspace{-16}$So $\bf{g\left(-\sqrt{2}\right)=2-\sqrt{3}\implies f\left(t\right)=-2+3\sqrt{2}\;}$ at $\bf{t=-\sqrt{2}}$\\\\\\ and $\bf{g\left(1-\sqrt{2}\right)=1-2\sqrt{2}\implies f\left(t\right)=-1+2\sqrt{2}}$ at $\bf{t=1-\sqrt{2}\;}$\\\\\\ and $\bf{g\left(\sqrt{2}\right)=2+3.\sqrt{2}\implies f\left(t\right)=2+3\sqrt{2}}$ at $\bf{t=\sqrt{2}}$\\\\\\ So $\boxed{\boxed{\bf{\min\left|\sin x+\cos x+\tan x+\cot x+\csc x+\sec x\right|=2\sqrt{2}-1}}}$\\\\\\ *We Can also use $\bf{\mathbb{A.M\geq G.M}}$ for Calculation of Min. of $\bf{f(t)=t+\frac{2}{t-1}=\left(t-1\right)+\frac{2}{t-1}}$

3√2+2

i think i misinterpreted....i took the modulus of the individual functions...i guess 2 is about right.