Find the range of f(x) = 1/|sin x| + 1/|cos x|

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fx=1sinx+1cosx=cosecx+secxLet us consider first quadrant, where sosec and sec are positivefx=cosecx+secxNow we know that maximum value of sec and cosec tend towards infinity.Hence we only need to find minimum value.Differentiating with respect to x we get:f'x=-cosecx.cotx+sec.tanxFor maixima or minima, f'x=0-cosecx.cotx+sec.tanx=0sec.tanx=cosecx.cotx1cosxcosxsinx=1sinx.sinxcosxsinx.cos2x=cosx.sin2xsinx.cos2x-cosx.sin2x=0sinx.cosxcosx-sinx=0When sinx=0, then cosec x is undefinedWhen cosx=0, then secx is undefinedHence these two cases are neglectedWhencosx-sinx=0sinx=cosxtanx=1x=π4There is no need to check for double derivative, as we know for sure that it will be pointof minimafπ4=cosecπ4+secπ4=2+2=22Hence range of fx is [22,)

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