The minimum value of the function f(x) = |sin | vedclass

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(D) Let $p = \sin x + \cos x$. Then $p^2 = 1 + 2 \sin x \cos x$,which implies $\sin x \cos x = \frac{p^2 - 1}{2}$.We know that $	an x + \cot x = \fra

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$2\sqrt{2} + 1$

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(D) Let $p = \sin x + \cos x$. Then $p^2 = 1 + 2 \sin x \cos x$,which implies $\sin x \cos x = \frac{p^2 - 1}{2}$.We know that $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{p^2 - 1}$.Also,$\sec x + \csc x = \frac{1}{\cos x} + \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x \cos x} = \frac{p}{\frac{p^2 - 1}{2}} = \frac{2p}{p^2 - 1}$.Substituting these into the function: $f(x) = |p + \frac{2}{p^2 - 1} + \frac{2p}{p^2 - 1}| = |p + \frac{2(1 + p)}{p^2 - 1}| = |p + \frac{2(1 + p)}{(p - 1)(p + 1)}| = |p + \frac{2}{p - 1}|$.We can rewrite this as $f(x) = |(p - 1) + \frac{2}{p - 1} + 1|$.For $p - 1 > 0$,by $AM$-$GM$ inequality,$(p - 1) + \frac{2}{p - 1} \geq 2\sqrt{(p - 1) \cdot \frac{2}{p - 1}} = 2\sqrt{2}$.Thus,$f(x) \geq 2\sqrt{2} + 1$.

The minimum value of the function $f(x) = |\sin x + \cos x + \tan x + \cot x + \sec x + \csc x|$ is equal to

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