Minimum value of the function f(x) = left| {s | vedclass

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Question

Put $=\sin x+\cos x=p \Rightarrow 1+2 \sin x \cos x=p^{2}$

$\sin x \cos x=\left(\frac{p^{2}-1}{2}ight)$

$	herefore f(x) = \left| {p + \frac{2

Vedclass · Accepted Answer

$2\sqrt 2  - 1$

Vedclass · Answer

Put $=\sin x+\cos x=p \Rightarrow 1+2 \sin x \cos x=p^{2}$

$\sin x \cos x=\left(\frac{p^{2}-1}{2}ight)$

$	herefore f(x) = \left| {p + \frac{2}{{\left( {{p^2} - 1} ight)}} + \frac{{2p}}{{{p^2} - 1}}} ight|$

${=\left|p+\frac{2}{(p-1)}ight|} $

${=\left|(p-1)+\frac{2}{(p-1)}+1ight|}$

$	herefore \frac{(p-1)+\left(\frac{2}{p-1}ight)}{2} \geq\left((p-1) \cdot \frac{2}{p-1}ight)^{1 / 2}$

$	herefore(p-1)+\left(\frac{2}{p-1}ight) \geq 2 \sqrt{2}$

$	herefore \quad f(x)_{\min } \geq 2 \sqrt{2}+1$

Minimum value of the function $f(x) = \left| {\sin \,x + \cos \,x + \tan \,x + \cot \,x + \sec \,x + \ cosec\ x} \right|$ is equal to

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