 Extreme values and Periodicity of Trigonometric functions
Question

# If the minimum value of $\mid \mathrm{sin}\mathrm{x}+\mathrm{cos}\mathrm{x}+\mathrm{tan}\mathrm{x}+\mathrm{cot}x+\mathrm{sec}\mathrm{x}+\mathrm{cosec}\mathrm{x}\mid ,\mathrm{x}\in \mathrm{R}$ is $\sqrt{\mathrm{m}}+\mathrm{n}$ then m + n equal $\left(\mathrm{m},\mathrm{n}\in \mathrm{z}\right)$

Difficult
Solution

## Playing a little bit with the expression one discovers that if we get ,then given expression $=\left|\mathrm{t}+\frac{1}{\frac{{\mathrm{t}}^{2}-1}{2}}+\frac{\mathrm{t}}{\frac{{\mathrm{t}}^{2}-1}{2}}\right|$$\begin{array}{l}=\left|\mathrm{t}+\frac{2}{{\mathrm{t}}^{2}-1}+\frac{2\mathrm{t}}{{\mathrm{t}}^{2}-1}\right|\\ \mathrm{y}=\left|\mathrm{t}+\frac{2}{\mathrm{t}-1}\right|=\left|\mathrm{t}-1+\frac{2}{\mathrm{t}-1}+1\right|\end{array}$If $1<\mathrm{t}\le \sqrt{2}$ then $\mathrm{y}=\mathrm{t}-1+\frac{2}{\mathrm{t}-1}+1\ge 2\sqrt{2}+1$If $-\sqrt{2}\le \mathrm{t}<-1$ then $\mathrm{y}=-1+1-\mathrm{t}+\frac{2}{1-\mathrm{t}}\ge 2\sqrt{2}-1$$\begin{array}{l}\therefore {\mathrm{y}}_{min}=2\sqrt{2}-1=\sqrt{8}-1\\ \therefore \mathrm{m}+\mathrm{n}=7\end{array}$

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